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1.1 Introduction .
In Class LX, you began your exploration of the world of real numbers and encountered
irrational numbers. We continue our discussion on real numbers in this chapter. We
begin with very important properties of positive integers in Sections 1.2, namely the
Euclid's division algorithm and the Fundamental Theorem of Arithmetic.

Euclid's division algorithm, as the name suggests, has to do with divisibility of
integers. Stated simply, it says any positive integer a can be divided by another positive
integer 5 in such a way that it leaves a remainder r that is smaller than b. Many of you
probably recognise this as the usual long division process. Although this result is quite
easy to state and understand, it has many applications related to the divisibility properties
of integers. We touch upon a few of them, and use it mainly to compute the HCF of
two positive integers. J) ;

The Fundamental Theorem of Arithmetic, on the other hand, has to do something
with multiplication of positive integers. You already know that every composite number
can be expressed as a product of primes in a unique way-this important fact is the
Fundamental Theorem of Arithmetic. Again, while it is a result that is easy to state and
understand, it has some very deep and significant applications in the field of mathematics.
We use the Fundamental Theorem of Arithmetic for two main applications. First, we
use it to prove the irrationality of many of the numbers you studied in Class IX, such as
V2, V3 and 5 . Second, we apply this theorem to explore when exactly the decimal
expansion of a rational number, say a #0), is terminating and when it is nonterminating repeating. We do so by looking at the prime factorisation of the denominator
qgof ? You will see that the prime factorisation of g will completely reveal the nature

q
of the decimal expansion of P,

q

So let us begin our exploration.

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2 MATHEMATICS
1.2. The Fundamental Theorem of Arithmetic

In your earlier classes, you have seen that any natural number can be written as a
product of its prime factors. For instance, 2 = 2,4 =2 x 2,253 = 11 x 23, and so on.
Now, let us try and look at natural numbers from the other direction. That is, can any
natural number be obtained by multiplying prime numbers? Let us see.

Take any collection of prime numbers, say 2, 3, 7, 11 and 23. If we multiply
some or all of these numbers, allowing them to repeat as many times as we wish,
we can produce a large collection of positive integers (In fact, infinitely many).
Let us list a few : :

7x 11x 23=1771 3x 7x 11 x 23 =5313

2x3x7x 11 x 23 = 10626 2? x 3 x 7? = 8232

2?x3x 7x 11 x 23 =21252 :
and so on. : ON

Now, let us suppose your collection of primes includes all the possible primes.
What is your guess about the size of this collection? Does it contain only a finite
number of integers, or infinitely many? Infact, there are infinitely many primes. So,
if we combine all these primes in all possible ways, we will get an infinite
collection of numbers, all
the primes and all possible ,
products of primes. The -
question is - can we .
produce all the composite ,
numbers this way? What "
do you think? Do you"
think that there may be a
composite number which
is not the product. of
powers of primes? >
Before we answer this,
let us factorise positive
integers, that is, do the
opposite of what we have
done so far. °

We are going to use
the factor tree with which
you are all familiar. Let us
take some large number,
say, 32760, and factorise 7 | 3 |
. 7 13
it as shown.

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REAL NUMBERS 3

So we have factorised 32760 as 2 x 2x 2x 3x3x5x7-x 13 as a product of
primes, i.e., 32760 = 23 x 3° x 5 x 7 x 13 as a product of powers of primes. Let us try
another number, say, 123456789. This can be written as 3° x 3803 x 3607. Of course,
you have to check that 3803 and 3607 are primes! (Try it out for several other natural
numbers yourself.) This leads us to a conjecture that every composite number can be
written as the product of powers of primes. In fact, this statement is true, and is called
the Fundamental Theorem of Arithmetic because of its basic crucial importance
to the study of integers. Let us now formally state this theorem.

Theorem 1.1 (Fundamental Theorem of Arithmetic) : Every composite
number can be expressed (factorised) as a product of primes, and this factorisation
is unique, apart from the order in which the prime factors occur.

An equivalent version of Theorem 1.2 was probably

first recorded as Proposition 14 of Book IX inEuclid's .)
Elements, before it came to be known as the | Qa ee
Fundamental Theorem of Arithmetic. However, the = Walt
first correct proof was given by Carl Friedrich Gauss SNe y
in his Disquisitiones Arithmeticae. _ ° ">
Carl Friedrich Gauss is often referred to as the 'Prince AY
of Mathematicians' and is considered one of the three \ SA
greatest mathematicians of all time, along with

Archimedes and Newton. He has made fundamental Carl Friedrich Gauss
contributions to both mathematics and science. (1777 - 1855)

The Fundamental Theorem of Arithmetic says that every composite number can
be factorised as a product of primes. Actually it says more. It says that given any
composite number it can be factorised as a product of prime numbers in a 'unique'
way, except for the order in which the primes occur. That is, given any composite
number there is one and only one way to write it as a product of primes, as long as we
are not particular about the order in which the primes occur. So, for example, we
regard 2 x 3 x 5 x 7 as the same as 3 x 5 x 7 x 2, or any other possible order in which
these primes are written. This fact is also stated in the following form:

The prime factorisation of a natural number is unique, except for the order
of its factors.

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4 MATHEMATICS
In general, given a composite number x, we factorise it as x = p,p, ... p,, where
Py» Py Pp, are primes and written in ascending order, i.e., p, < p,
<...<p_. If we combine the same primes, we will get powers of primes. For example,
32760 =2x2x2x3x3x5x7x13=2?x3?x5x7x 13
Once we have decided that the order will be ascending, then the way the number
is factorised, is unique.
The Fundamental Theorem of Arithmetic has many applications, both within
mathematics and in other fields. Let us look at some examples.
Example 1 : Consider the numbers 4", where n is a natural number. Check whether
there is any value of n for which 4" ends with the digit zero.
Solution : If the number 4", for any n, were to end with the digit zero, then it would be
divisible by 5. That is, the prime factorisation of 4" would contain the prime 5. This is
not possible because 4" = (2)""; so the only prime in the factorisation of 4" is 2. So, the
uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no
other primes in the factorisation of 4". So, there is no natural number n for which 4"
ends with the digit zero. we -
You have already learnt how to find the HCF and LCM of two positive integers
using the Fundamental Theorem of Arithmetic in earlier classes, without realising it!
This method is also called the prime factorisation method. Let us recall this method
through an example. - _
fo

Example 2 : Find the LCM and HCF of 6 and 20 by the prime factorisation method.
Se" .

Solution: We have: 6=2'x 3! and 20=2x2x5=2?x5!,

You can find HCF(6, 20) = 2 and LCM(6, 20) = 2 x 2 x 3 x 5 = 60, as done in your

earlier classes.

Note that HCF(6, 20) = 2' = Product of the smallest power of each common

oo * prime factor in the numbers.
LCM (6, 20) = 2? x 3' x 5' = Product of the greatest power of each prime factor,
involved in the numbers.

From the example above, you might have noticed that HCF(6, 20) x LCM(6, 20)
= 6 x 20. In fact, we can verify that for any two positive integers a and b,
HCF (a, b) x LCM (a, b) =a x b. We can use this result to find the LCM of two
positive integers, if we have already found the HCF of the two positive integers.
Example 3: Find the HCF of 96 and 404 by the prime factorisation method. Hence,
find their LCM.

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REAL NUMBERS 5
Solution : The prime factorisation of 96 and 404 gives :
96 = 2° x 3, 404 =2°x 101
Therefore, the HCF of these two integers is 2? = 4.
96 x 404 96 x 404
Also, LCM (96, 404) = --__--_ = ---- = 9696
HCF(96, 404) 4
Example 4 : Find the HCF and LCM of 6, 72 and 120, using the prime factorisation
method. "
Solution : We have :
6=2x3, 72=2?x 37, 120=2?x3x5
Here, 2' and 3! are the smallest powers of the common factors 2 and 3, respectively.
So, HCF (6, 72, 120)= 2'x3'=2x3=6. 0 7
23, 3? and 5! are the greatest powers of the prime factors 2, 3 and 5 respectively
involved in the three numbers. oO oO -
So, LCM (6, 72, 120) = 23x 3? x 5' = 360
Remark : Notice, 6 x 72 x 120 # HCF (6, 72, 120) x LCM (6, 72, 120). So, the
product of three numbers is not equal to the product of their HCF and LCM.
C) EXERCISE 1.1
NO .
1. Express each number as a product of its prime factors:
@) 140 (ii) 156 (iit) 3825 (tv) 5005 (v) 7429
2. Find the LCM and HCF of the following pairs of integers and verify that LCM x HCF =
product of the two numbers.
@) 26and91 ~ Gi) 510 and 92 (iii) 336 and 54
3. Find the LCM and HCF of the following integers by applying the prime factorisation
method.
@) 12,15 and 21 (ii) 17,23 and 29 (iii) 8,9 and 25
4, Given that HCF (306, 657) = 9, find LCM (306, 657).
5. Check whether 6" can end with the digit 0 for any natural number n.
6. Explain why 7 x 11x 13+13and7x6x5x4x3x2x1+5 are composite numbers.
7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round
of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the
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6 MATHEMATICS
same point and at the same time, and go in the same direction. After how many minutes
will they meet again at the starting point?

1.3 Revisiting Irrational Numbers

In Class IX, you were introduced to irrational numbers and many of their properties.

You studied about their existence and how the rationals and the irrationals together

made up the real numbers. You even studied how to locate irrationals on the number

line. However, we did not prove that they were irrationals. In this section, we will
prove that /2 , V3 , 5 and, in general, Jp is irrational, where p is a prime. One of
the theorems, we use in our proof, is the Fundamental Theorem of Arithmetic.

Recall, a number 's' is called irrational if it cannot be written in the form 7
where p and gq are integers and g + 0. Some examples of irrational numbers, with
which you are already familiar, are : . :

2.8, M15, x, -2F> 0.101101 HOLD...

Before we prove that ./2 is irrational, we need the following theorem, whose
proof is based on the Fundamental Theorem of Arithmetic.

Theorem 1.2 : Let p be a prime number. If p divides a', then p divides a, where

a is a positive integer. .

*Proof : Let the prime factorisation of a be as follows :

t +] ,
a=pp,...P,, where p,,p,,..., P, are primes, not necessarily distinct.

Therefore, a? = (p,p,.--P,)(P\P, +++ P,) = PPS + + + Pe

Now, we are given that p divides a'. Therefore, from the Fundamental Theorem of

Arithmetic, it follows that p is one of the prime factors of a*. However, using the

uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only

prime factors of a' are p,, p,, . . -, P,. So p is one of p,, Py... » P,:

Now, since a= p, p,...P,, p divides a.

We are now ready to give a proof that ,/2 is irrational.

The proof is based on a technique called 'proof by contradiction'. (This technique is

discussed in some detail in Appendix 1).

Theorem 1.3: ./2 is irrational.

Proof : Let us assume, to the contrary, that ./2 is rational.

* Not from the examination point of view.

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REAL NUMBERS 7
r
So, we can find integers r and s (# 0) such that /2 = S°
Suppose r and s have a common factor other than 1. Then, we divide by the common
factor to get [2 =", where a and b are coprime.
b
So, b./2 =a.
Squaring on both sides and rearranging, we get 2b" = a'. Therefore, 2 divides a'.
Now, by Theorem 1.2, it follows that 2 divides a.
So, we can write a = 2c for some integer c. :
Substituting for a, we get 2b' = 4c', that is, b? = 2c'. .
This means that 2 divides 5', and so 2 divides b (again using Theorem 1.2 with p = 2).
Therefore, a and } have at least 2 as a common factor. .
But this contradicts the fact that a and b have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that ./2 is rational.
So, we conclude that ./2 is irrational. a _
Example 5 : Prove that ./3 is irrational.
Solution : Let us assume, to the contrary, that /3_ is rational.
oN a
That is, we can find integers a and b (# 0) such that /3 = b
Se" .

Suppose a and 5b have a common factor other than 1, then we can divide by the
common factor, and assume that a and b are coprime.
So, bV3 =a: . .
Squaring on both sides, and rearranging, we get 3b? = a'.
Therefore, a' is divisible by 3, and by Theorem 1.2, it follows that a is also divisible
by 3.
So, we can write a = 3c for some integer c.
Substituting for a, we get 3b' = 9c', that is, b? = 3c'.
This means that 5? is divisible by 3, and so 5 is also divisible by 3 (using Theorem 1.2
with p = 3).

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8 MATHEMATICS
Therefore, a and b have at least 3 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that ./3 is rational.
So, we conclude that ./3 is irrational.
In Class IX, we mentioned that :
e@ the sum or difference of a rational and an irrational number is irrational and
e@ the product and quotient of a non-zero rational and irrational number is
irrational,
We prove some particular cases here. .
Example 6 : Show that 5 - J3 is irrational. . .
Solution : Let us assume, to the contrary, that 5 - J3 is rational. .
. . . : oe a
That is, we can find coprime a and b (b #0) such that 5 - V3 = b
Therefore, 5 - " = J3- .
b .
Rearranging this equation, we get /3 =5 -" = b= a
- ob b
. aan a. . . .
Since a and b are integers, we get 5 - Db is rational, and so 3 is rational.
yf
But this contradicts the fact that ./3 is irrational.
This contradiction has arisen because of our incorrect assumption that 5 - V3 is
rational.
So, we conclude that 5 - J3 is irrational.
Example 7 : Show that 3V2 is irrational.
Solution : Let us assume, to the contrary, that 3/2. is rational.
. "on . a
That is, we can find coprime a and b (b # 0) such that 3/2 = D
Rearranging, we get 2 =".
3b
. . a, . . .
Since 3, a and b are integers, ap |S rational, and so ,/2 is rational.
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REAL NUMBERS 9
But this contradicts the fact that ./2 is irrational.
So, we conclude that 3./2 is irrational.
EXERCISE 1.2
1. Prove that ./5 is irrational.
2. Prove that 3+ 2/5 is irrational.
3. Prove that the following are irrationals : .
1 :
® = Gi) 7/5 (iii) 6+ V2 |
i v5 .
1.4 Summary : CO
In this chapter, you have studied the following points: . ,
1. The Fundamental Theorem of Arithmetic : , °
Every composite number can be expressed (factorised) as a product of primes, and this
factorisation is unique, apart from the order in which the prime factors occur.
2. If pis a prime and p divides a', then p divides a, where a is a positive integer.
3. To prove that /2, J3 are irrationals. . :
f° . .
A NOTE TO THE READER
You have seen that :
HCE (p, g, r) x LCM (p, g, r) #p x q Xr, where p, q, r are positive integers
(see Example 8). However, the following results hold good for three numbers
P,qandr:
-q:r- HCF(p, q, r
LOM (p, q.1) = pow PE a)
HCF( p, gq) - HCF(q,r) - HCF(p,r)
-q:-r LCM(p, q, r
HCF (p, q.) = Pq (P, 41)
LCM(p, q)-LCM(q, r)- LCM(p, r)
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Ia
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[a] S's '
1O62CHO2
2.1 Introduction .
In Class IX, you have studied polynomials in one variable and their degrees. Recall
that if p(x) is a polynomial in x, the highest power of x in p(x) is called the degree of
the polynomial p(x). For example, 4x + 2 is a polynomial in the variable x of
degree 1, 2y?- 3y + 4 is a polynomial in the variable y of degree 2, 5x3 -- 4x? +x- ./2
3
is a polynomial in the variable x of degree 3 and 7u°- at +4u*? +u-B8isa polynomial
eel 1
in the variable u of degree 6. Expressions like --, /x +2, =. ete., are
. ~x-1 x" +2x+3
not polynomials. No ) .

A polynomial of degree 1 is called a linear polynomial. For example, 2x - 3,
V3x +5, yt V2 ,x- =, 3z+4, 5u +1, etc., are all linear polynomials. Polynomials
such as 2x + 5 - x', x3 + |, etc., are not linear polynomials.

A polynomial of degree 2 is called a quadratic polynomial. The name 'quadratic'
has been derived from the word 'quadrate', which means 'square'. 2x* + 3x - =
y? - 2, 2- x7 + V3x, 37 Qu? +5, Sy? - 3" 47? +2 are some examples of
quadratic polynomials (whose coefficients are real numbers). More generally, any
quadratic polynomial in x is of the form ax? + bx + c, where a, b, c are real numbers
and a #0. A polynomial of degree 3 is called a cubic polynomial. Some examples of

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POLYNOMIALS 11
a cubic polynomial are 2 - x3, x3, 2°, 3-2? +. x3, 3x3- 2x2 + x- 1. In fact, the most
general form of a cubic polynomial is

axe + bx? + cx +d,
where, a, b, c, d are real numbers and a # 0.

Now consider the polynomial p(x) = x? - 3x - 4. Then, putting x = 2 in the
polynomial, we get p(2) =2?- 3 x 2-4 =- 6. The value '- 6', obtained by replacing
x by 2 in x? - 3x - 4, is the value of x? - 3x - 4 at x = 2. Similarly, p(0) is the value of
P(x) at x = 0, which is - 4. :

If p(x) is a polynomial in x, and ifk is any real number, then the value obtained by
replacing x by k in p(x), is called the value of p(x) at x =k, and is denoted by p(k).
What is the value of p(x) = x? -3x-- 4 at x=-1? We have: . :

p(-l) = C1P-{3 x CD} -4=0 So
Also, note that p(4) = 4?-(3 x 4)-4=0. ° . oe

As p(-1) = 0 and p(4) = 0, -1 and 4 are called the zeroes of the quadratic
polynomial x* - 3x - 4. More generally, a real number & is said to be a zero of a
polynomial p(x), if p(k) = 0. . .

You have already studied in Class IX, how to find the zeroes of a linear
polynomial. For example, if k is a zero of p(x) = 2x + 3, then p(k) = 0 gives us

3 . 5
2k+3=0,10¢.,k= 57° "
(2 . -b

In general, if k is a zero of p(x) = ax + b, then p(k) = ak + b=0,ie., k= a
So, the zero of the linear polynomial ax + b is > = = (Constant term) .

a Coefficient of x

Thus, the zero of a linear polynomial is related to its coefficients. Does this
happen in the case of other polynomials too? For example, are the zeroes of a quadratic
polynomial also related to its coefficients?

In this chapter, we will try to answer these questions. We will also study the
division algorithm for polynomials.

2.2 Geometrical Meaning of the Zeroes of a Polynomial
You Know that a real number k is a zero of the polynomial p(x) if p(k) = 0. But why
are the zeroes of a polynomial so important? To answer this, first we will see the
geometrical representations of linear and quadratic polynomials and the geometrical
meaning of their zeroes.

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12 MATHEMATICS
Consider first a linear polynomial ax + b, a #0. You have studied in Class IX that the
graph of y = ax + bis a straight line. For example, the graph of y = 2x + 3 is a straight
line passing through the points (- 2, -1) and (2, 7).
ee
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between x =-1and.x=-2, 9 Seay
that is, at the point (-3, 0). 7
2x+3is-=. Thus, the zeroof Go) yey
2 SEE Ree nner aera areca Vaan rere rece eee
the polynomial 2x + 3 is the - SgesRsSe RSS ER es tet aes Hest ei peaiie
x-coordinate of the point where the -
graph of y = 2x + 3 intersects the Fig. 2.1
: f- ;
x-axis, )
NO .
In general, for a linear polynomial ax + b, a # 0, the graph of y=ax+bisa
. ; wa: . . -b
straight line which intersects the x-axis at exactly one point, namely, | -.0].
. * a
Therefore, the linear polynomial ax + b, a # 0, has exactly one zero, namely, the
x-coordinate of the point where the graph of y = ax + b intersects the x-axis.

Now, let us look for the geometrical meaning of a zero of a quadratic polynomial.
Consider the quadratic polynomial x* - 3x - 4. Let us see what the graph* of
y =x' - 3x -4 looks like. Let us list a few values of y = x* - 3x - 4 corresponding to
a few values for x as given in Table 2.1.

* Plotting of graphs of quadratic or cubic polynomials is not meant to be done by the students,
nor is to be evaluated.
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POLYNOMIALS 13
Table 2.1

fe fa fof a fa fs [5

pressed {io | oo [a4] -6 [-o | -4 | 0 [6 |

If we locate the points listed So EE TT Te Ee
the graph, it will actually look like og oo

. . ooo oo fee.
upwards like \J or open | oo
downwards like {\ depending on ee
curves are called parabolas.) sve ce
quadratic_polynomial. ) Also to
are the x-coordinates of the points HARSHEST Tg A ee
where the graph of y = x* - 3x-4 - |.
intersects the x-axis. Thus, the ||)
x° - 3x - 4 are x-coordinates of HAA ee
the points where the graph of Fie. 2.2
y = x? - 3x - 4 intersects the 1B. *
x-axis. .

This fact is true for any quadratic polynomial, i.e., the zeroes of a quadratic
polynomial ax? + bx +c, a #0, are precisely the x-coordinates of the points where the
parabola representing y = ax' + bx + c intersects the x-axis.

From our observation earlier about the shape of the graph of y = ax? + bx +c, the
following three cases can happen:

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14 MATHEMATICS
Case (i) : Here, the graph cuts x-axis at two distinct points A and A'.
The x-coordinates of A and A' are the two zeroes of the quadratic polynomial
ax? + bx + c in this case (see Fig. 2.3).
Sa Ee Earner | Hee eee eS Ge ere aie ee eee ie an | pe a Ee RE
Sa aE earn Hea | HEY EE Ge ere aie ee eee ie an | ApS ER
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eae eee ete | oF /ae eee ee bees Pape teeth Steere | eee fate
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(i) . oO, (it)
Fig. 2.3 ° :
Case (ii) : Here, the graph cuts the x-axis at exactly one point, i.e., at two coincident
points. So, the two points A and A' of Case (i) coincide here to become one point A
(see Fig. 2.4). . .
Pe Pe es Peete trie treet tee tty piece tee rites Piet tat
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(i) (ii)
Fig. 2.4
The x-coordinate of A is the only zero for the quadratic polynomial ax'? + bx + c
in this case.
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POLYNOMIALS 15
Case (iii) : Here, the graph is either completely above the x-axis or completely below
the x-axis. So, it does not cut the x-axis at any point (see Fig. 2.5).
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fof hdr"
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RECESSES EEE CEE EEE EE EEE Ee Se Soe ENA CHEE EH
oo | of
(i) . (ii)
, Fig. 2.5.
So, the quadratic polynomial ax? + bx + c has no zero in this case.

So, you can see geometrically that a quadratic polynomial can have either two
distinct zeroes or two equal zeroes (i.¢e., one zero), or no zero. This also means that a
polynomial of degree 2 has atmost two zeroes.

Now, what do you expect the geometrical meaning of the zeroes of a cubic
polynomial to be? Let us find out. Consider the cubic polynomial x* - 4x. To see what
the graph of y = x* - 4x looks like, let us list a few values of y corresponding to a few
values for x as shown in Table 2.2.

os Table 2.2
fe ta fo fo fe
y=x-4x 0 3 0 -3 0

Locating the points of the table on a graph paper and drawing the graph, we see

that the graph of y = x° - 4x actually looks like the one given in Fig. 2.6.
Reprint 2025-26


16 MATHEMATICS
cubic polynomial x° - 4x. Observe gi ee ssa ie Nii ee
intersects the x-axis. Since the curve Se

in Fig. 2.7 and Fig. 2.8 respectively. °

oe - Fig. 2.6
ee
SUL ob cit fF.
| ey ee
Net ae pepo BA eet a
ee ee ee
et eels /e2 oo ei ee) L Se

Fig. 2.7 Fig. 2.8

Reprint 2025-26


POLYNOMIALS 17

Note that 0 is the only zero of the polynomial x*. Also, from Fig. 2.7, you can see
that 0 is the x-coordinate of the only point where the graph of y = x? intersects the
x-axis. Similarly, since x*--x* =x°(x- 1), 0 and 1 are the only zeroes of the polynomial
x* - x°. Also, from Fig. 2.8, these values are the x-coordinates of the only points
where the graph of y = x* - x° intersects the x-axis.

From the examples above, we see that there are at most 3 zeroes for any cubic
polynomial. In other words, any polynomial of degree 3 can have at most three zeroes.
Remark : In general, given a polynomial p(x) of degree n, the graph of y = p(x)
intersects the x-axis at atmost m points. Therefore, a polynomial p(x) of degree n has
at most n zeroes.

Example 1 : Look at the graphs in Fig. 2.9 given below. Each is the graph of y = p(x),
where p(x) is a polynomial. For each of the graphs, find the number of zeroes of p(x).
EEE ERE EES EEE EE pect EER EEE SRE SRE REE EE ESS og es PE ey A SP fa eat ot | ated Heed cece Heath. HH
ee ee
ee
SET E/E Ey GSTain SEs baie ala
SS Se z (CSTE le
Og yve EE Be ve ve
) oY (ii)
ee
Naps, Mega P eps NA
Ce ee
ee ee
Pye bE vee ee yea
(iv) (v) (vi)
, . Fig. 2.9
Solution :

(i) The number of zeroes is | as the graph intersects the x-axis at one point only.
(ii) The number of zeroes is 2 as the graph intersects the x-axis at two points.
(ii) The number of zeroes is 3. (Why?)

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18 MATHEMATICS
(iv) The number of zeroes is 1. (Why?)
(v) The number of zeroes is 1. (Why?)
(vi) The number of zeroes is 4. (Why?)
EXERCISE 2.1
1. The graphs of y = p() are given in Fig. 2.10 below, for some polynomials p(x). Find the
number of zeroes of p(x), in each case.
pete et ete
Se ee ee ee e/a
BU Gee eee ee GE aN eae
ae dee O Ee re arp a a 20 (aia # ay iO sd ey <
CO ON
plsbsehe Patel Pisigisie| nciaedd Bin detaie sistas
A ye
(i) 7 (iii)
Pte te et
Spee ee ye ee ES
XG NWEVASe SEs - ae e ap (sa ea YAING i
Cae ee
eitistets|etaicten| Pietersie airtete isiersie| eeirhty
ee eye ye
(iv) (v) (vi)
' Fig. 2.10
2.3 Relationship between Zeroes and Coefficients of a Polynomial
You have already seen that zero of a linear polynomial ax + bis - b . We will now try
a
to answer the question raised in Section 2.1 regarding the relationship between zeroes
and coefficients of a quadratic polynomial. For this, let us take a quadratic polynomial,
say p(x) = 2x? - 8x + 6. In Class IX, you have learnt how to factorise quadratic
polynomials by splitting the middle term. So, here we need to split the middle term
"- 8x' as a sum of two terms, whose product is 6 x 2x? = 12x'. So, we write
2x* - 8x + 6 = 2x? - 6x - 2x + 6 = 2x(x - 3) - 2(x - 3)
= (2x - 2)(x - 3) = 2(4 - 1)(x - 3)
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POLYNOMIALS 19
So, the value of p(x) = 2x° - 8x + 6 is zero when x- 1 =0 orx-3 =0, i.e., when
x = 1 orx = 3. So, the zeroes of 2x° - 8x + 6 are 1 and 3. Observe that :
Sum of its zeroes = 1+3=4= -C) = -Coetficient of x) of )
2 Coefficient of x
Product of its zeroes = 1x3 =3 = 6 = _Constant term _
2 Coefficient of x
Let us take one more quadratic polynomial, say, p(x) = 3x° + 5x - 2. By the
method of splitting the middle term, "
3x° + 5x-2= 3x? 4+ 6x--x-2=3x(%4+2)-l(¥4+2) .
= 3x - L@ + 2) os
Hence, the value of 3x? + 5x - 2 is zero when either 3x -1=Oo0rx+2=0,ie.,
1 1 +
when x = 3 or x =-2. So, the zeroes of 3x? + 5x - 2 are 3 and - 2. Observe that :
5 fit 1 ( 2) -5 -(Coefficient of x)
um of its zeroes > = [t+ (--2) =-- = - ese
3 ~ 3°. Coefficient of x7
| 1°) =2 Constant t
Product of its zeroes = - X (-2) =-- = eee
. 3 ~ -3 Coefficient of x
_
In general, if «* and B* are the zeroes of the quadratic polynomial p(x) = ax' + bx +c,
a #0, then you know that x - a and x - B are the factors of p(x). Therefore,
ax? + bx + c = k(x - &) (x - B), where k is a constant
= k[x? -(a + B)x + a B)
_- =k' --ka+pBx+kaB
Comparing the coefficients of x', x and constant terms on both the sides, we get
. . a=k,b=-k(a +) and c = kab.
_ -b
This gives a+B= a?
c
ap = -
B a
* a,B are Greek letters pronounced as 'alpha' and 'beta' respectively. We will use later one
more letter "y' pronounced as 'gamma'.
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20 MATHEMATICS
. b - -(Coefficient of x)
Le., sum of zeroes = a + B = --- = --__-_____,
a Coefficient of x
product of zeroes = af = - _Constant term
a Coefficient of x
Let us consider some examples.
Example 2 : Find the zeroes of the quadratic polynomial x? + 7x + 10, and verify the
relationship between the zeroes and the coefficients.
Solution : We have
+ 7x + 10 =(x + 2)(x +5) /
So, the value of x? + 7x + 10 is zero when x + 2 =0 orx+5=0,i.e., when x =-2 or
x =-5, Therefore, the zeroes of x7 + 7x + 10 are - 2 and - 5. Now, ,
" - 7 . -(Coefficient of x
sum of zeroes = -2 + (-5)=-(7) = 0 = Coefficient of 9),
; . 1 - Coefficient of x
. 10 Constant t
product of zeroes = (-2) x (-5) =10=-= --rev
d Coefficient of x
Example 3 : Find the zeroes of the polynomial x? - 3 and verify the relationship
between the zeroes and the coefficients.
Solution : Recall the identity a? - b? = (a - b)(a + b). Using it, we can write:
eae (x-WB)(0+ 98)

So, the value of x' - 3 is zero when x = J3 orx = -¥3Therefore, the zeroes of x? - 3 are J3 and -/3.
Now, |

sum of zeroes = V3-V3= 0= Coefficient of x)

Coefficient of x
-3 Constant term
duet of zeroes = (v/3)(-V3) = -3 = 3 = Constant term _
PROMUCE OF ZPOCS ( ) ) 1 Coefficient of x'
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POLYNOMIALS 21
Example 4 : Find a quadratic polynomial, the sum and product of whose zeroes are
-3 and 2, respectively.
Solution : Let the quadratic polynomial be ax" + bx + c, and its zeroes be o and B.
We have
B=-3=-
a+Bp=-3=-:
a
d B=2-<
ap= = °

an a
If a=1, then b=3 andc=2., ,
So, one quadratic polynomial which fits the given conditions is x? +3x4+2. ©

You can check that any other quadratic polynomial that fits these conditions will
be of the form k(x? + 3x + 2), where k is real. oe

Let us now look at cubic polynomials. Do you think a similar relation holds
between the zeroes of a cubic polynomial and its coefficients? ~

Let us consider p(x) = 2x3-5x2-14x4+8. .

. log.

You can check that p(x) = 0 for x = 4, - 2, 5 Since p(x) can have atmost three

zeroes, these are the zeores of 2x? - 5x - 14x + 8. Now,
CN 1° 5 _ -(-5) _ -(Coefficient of x?
sum of the zeroes = 4+ (-2)¢ +52 =O) - Coefficient of)
: (2 2 2 Coefficient of x
product of the zeroes = 4 x (-2) x _ -4= a -Constant term
2 2 Coefficient of x

However, there is one more relationship here. Consider the sum of the products
of the zeroes taken two at a time. We have

{4x (-2)}+ (-2) x4 + ta

2 2
_ 91422-7224 _ Coefficient of x .
2 Coefficient of x

In general, it can be proved that if o, B, y are the zeroes of the cubic polynomial

axe + bx? + cx + d, then
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22 MATHEMATICS
-b
a+rB+y= a"
c
op + By + yo = a?
-d
aBp y= a
Let us consider an example.
; 1 ; ,
Example 5* : Verify that 3, -1, "3 are the zeroes of the cubic polynomial
p(x) = 3x3 - 52x? - 11x -3, and then verify the relationship between the zeroes and the
coefficients. -
Solution : Comparing the given polynomial with ax? + bx? + cx +d, we get
a=3,b=-5,c=-ll,d=-3.Furthr en
p(3) =3 x 3°-(S x ) - (11 x 3) -3 = 81 - 45 - 33-3 =0,
p(-l) =3 x 1-5 x C1)?- 11 x (1) -3 =-3-5+11-3=0,
3 2°
emo
3 3 3 3
_ 1353 242-9
9.9 +3 3. 3
SS Le" .
1 .
Therefore, 3, -1 and "3 are the zeroes of 3x7 - 5x? - 11x - 3.
el
So, we take a = 3, B =-1 and y= "3
Now, a ;
oo 1 1 5 -€5) --b
o+B+ y=34+(-D)+| -=}=2--=-=---=- >
Bry o) ( ;] 3 3 3 a
1 1 1 -ll_c¢
aBb+ By +ya=3x(-D+C1x/-=]4+]-=|x3=-34+--l=-=-->
B+ By +y co+enx(-3}+(-3) 3 3 a
1 -(-3) _-d
aby =3x(-l) x| -- |=1l=--~ =-_By (-) ( | 3 ,
* Not from the examination point of view.
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POLYNOMIALS 23
EXERCISE 2.2
1. Find the zeroes of the following quadratic polynomials and verify the relationship between
the zeroes and the coefficients.
(i) x?-2x-8 (ii) 457-4541 (iii) 6x? -3 -7x
(iv) 4u?+8u (vy) P-15 (vi) 3x°-x-4
2. Find a quadratic polynomial each with the given numbers as the sum and product of its
zeroes respectively.
1 1 wt
@ orl ji) V2.5 Gi) OVS
iv) 1,1 it i) 41
(iv) 1, W) "7a (vi) 4, -
2.4 Summary . ;
In this chapter, you have studied the following points: a,
1. Polynomials of degrees 1, 2 and 3 are called linear, quadratic and cubic polynomials
respectively. CO "
2. A quadratic polynomial in x with real coefficients is of the form ax? + bx +c, where a, b,c
are real numbers with a #0. : ,
3. The zeroes of a polynomial p(x) are precisely the x-coordinates of the points, where the
graph of y = p(x) intersects the x-axis. | .
4, A quadratic polynomial can have at most 2 zeroes and a cubic polynomial can have
at most 3 zeroes.' :) :
5. If a and B are the zeroes of the quadratic polynomial ax? + bx + c, then
a+p=-2, op =<
a a
6. Ifa, B, yare the zeroes of the cubic polynomial ax? + bx? + cx + d, then
a+B+y= ,
a
: c
aB+By+ ya=->
° a
-d
and apy =--
a
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Eleaf |G
La] EE os
viata A a
ieee
1062CHO3
3.1 Introduction -
You must have come across situations like the one given below: ©

Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel
and play Hoopla (a game in which you throw a ring on the items kept in a stall, and if
the ring covers any object completely, you get it). The number of times she played
Hoopla is half the number of rides she had on the Giant Wheel. If each ride costs
= 3, and a game of Hoopla costs = 4, how would you find out the number of rides she
had and how many times she played Hoopla, provided she spent ® 20.

May be you will try it by considering different cases. If she has one ride, is it
possible? Is it possible to have two rides? And so on. Or you may use the knowledge
of Class IX, to represent 'such situations as linear equations in two variables.

7 Tas DC
sila bees Tilo gf WG om ak ZN
ie Rea |
See eS NEN TN Oe
=- Gy Sy ' Xx
~ ES oy => ott, VI
ey \ N tommy Of ck Ta

} y eesti p if ILENE

gb BON =
: Ky Zh mp AY Dy = -a NY
ay) / 65 Ce " ~
ny? 1, SS AC"
= \e@ ' i | . . aes =~
St ihs 7 fl =f

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Pair OF LINEAR EQUATIONS IN TWo VARIABLES 25
Let us try this approach.
Denote the number of rides that Akhila had by x, and the number of times she
played Hoopla by y. Now the situation can be represented by the two equations:
< (1)
3x +4y = 20 (2)
Can we find the solutions of this pair of equations? There are several ways of
finding these, which we will study in this chapter. .
3.2. Graphical Method of Solution of a Pair of Linear Equations
A pair of linear equations which has no solution, is called an inconsistent pair of linear
equations. A pair of linear equations in two variables, which has a solution, is called a
consistent pair of linear equations. A pair of linear equations which are equivalent has
infinitely many distinct common solutions. Such a pair is called a dependent pair of
linear equations in two variables. Note that a dependent pair of linear equations is
always consistent. So . :
We can now summarise the behaviour of lines representing a pair of linear equations
in two variables and the existence of solutions as follows:
(i) the lines may intersect in a single point. In this case, the pair of equations
has a unique solution (consistent pair of equations).
So
(ii) the lines may be parallel. In this case, the equations have no solution
(inconsistent pair of equations).
(ii) the lines may be coincident. In this case, the equations have infinitely many
solutions [dependent (consistent) pair of equations].
Consider the following three pairs of equations.
(i) x--2y =0 and 3x+4y-20=0 (The lines intersect)
(ii) 2x+3y-9=Oand4x+6y--18=0 (The lines coincide)
(it) x + 2y-4=Oand 2x+4y-12=0 - (The lines are parallel)
. a, Bb, CQ,
Let us now write down, and compare, the values of & bh and > in all the
, 2 2 2
three examples. Here, a,, b,, c, and a,, b,, c, denote the coefficents of equations
given in the general form in Section 3.2.
Reprint 2025-26


26 MATHEMATICS
Table 3.1
a, b, q
SI | Pair of lines TO Compare the | Graphical Algebraic
ay b, | . . |. .
No. ratios representation interpretation
1. |x-2y=0 1} 2) 0 ja .4 I ing |Exactl
. |x-2y= 5 - | ay ntersecting |Exactly one
3x4 4y-20=0 3 4} -20|% lines solution
(unique)
> frreay-920 | 2] 2] [2 =2%="|coincigent -fintinite
. X+ Sy-F7= 4 6 -18 a, b, C5 lines ent niinite y .
ines many solutions
Ax + 6y-18=0
3 |xe2y-4-0 | + | 2 |S GGT nettines |N luti
. | X+ 4 _ - ~~ ~ aralle! lines 0 solution
y 2 4 -12|}% h &
2x+4y-12=0
From the table above, you can observe that if the lines represented by the equation
ax+by+c,=90 :
and axt+by+c,=0 :
. , a bh ;
are (i) intersecting, then -- ¥ --
i :) ay b, :
ey a ob -G
(ii) coincident, then - = - = -.
a, bo &
ase a _ bh G
(iii) parallel, then --=- 4 --
a a by &
In fact, the converse is also true for any pair of lines. You can verify them by
considering some more examples by yourself.
Let us now consider some more examples to illustrate it.
Example I : Check graphically whether the pair of equations
° x+3y=6 (1)
and 2x -3y= 12 (2)
is consistent. If so, solve them graphically.
Solution : Let us draw the graphs of the Equations (1) and (2). For this, we find two
solutions of each of the equations, which are given in Table 3.2
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Pair OF LINEAR EQUATIONS IN TWo VARIABLES 27
Table 3.2
pos foto] |x fof a
. Ee SEES SE cet EE SSE
Plot the points A(O,2), B(6, 0), a -
PO, - 4) and Q(3,- 2) on graph IA)
aper, and join the points toformthe {82 3.) [aa EG eee page
Hines AB and PQ as shown in a
Fie hu eee
ig. 3.1. eceaerieri tee eed ©) pee en Erie iee Berita ito ME eet i
 comemecnemm Lb
e observe that there is a point Sys H sys | RE Sey inte RR RTA RHE
point OG ay
B (6, 0) common to both the lines #:3:3:3s33:1 5) [sess Ss aN HE SESE SoSH
; Haceenr ey ae SAEs sated setae yan asc
AB and PQ. So, the solution of the | Be
. ; . Scere Sea HSE gE a re HH SE SH SESH SEH HSE
pair of linear equations isx=6and #2 Syke gj
y=0, i.e., the given pair of equations ed so ee
s consi ae
1s consistent. EEE SESH SE SG a eed USS SG SSE SSE EHH EH
Fig. 3.1
Example 2 : Graphically, find whether the following pair of equations has no solution,
unique solution or infinitely many solutions: - -
"- 5x-8y +1=0 (1)
3 24 3 0 (2)
-_- +- =
Ms TS
Solution : Multiplying Equation (2) by 3" we get
_. 5x=8y+1=0
But, this is the same as Equation (1). Hence the lines represented by Equations (1)
and (2) are coincident. Therefore, Equations (1) and (2) have infinitely many solutions.
Plot few points on the graph and verify it yourself.
Example 3 : Champa went to a 'Sale' to purchase some pants and skirts. When her
friends asked her how many of each she had bought, she answered, "The number of
skirts is two less than twice the number of pants purchased. Also, the number of skirts
is four less than four times the number of pants purchased". Help her friends to find
how many pants and skirts Champa bought.
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28 MATHEMATICS
Solution : Let us denote the number of pants by x and the number of skirts by y. Then
the equations formed are :
y= 2x-2 (1)
and y= 4x-4 (2)
Let us draw the grap hs of PERRET EERE EERE EEE EEE EEE E EE EERE EEE EEE EE EEE EEE ep eee
They are given in Table 3.3. ee
Table 3.3 SEER Peal SRE Pee Sai Ena ina
EE eet th') [HEH he tee eae ae
Oe, a.
ure erent PERE eee SEE EERE EEE
4 Sees eee eae ETE ATE ET
Hyede-4]-4] 0 | Saeed SeceE See Cede Le Eee Ee EEE HELE PELE
_- Ee
_ Fig. 3.2
Plot the points and draw the lines passing through them to represent the equations,
as shown in Fig. 3.2. \ )
The two lines intersect at the point (1, 0). So, x = 1, y = 0 is the required solution
of the pair of linear equations, i.e., the number of pants she purchased is 1 and she did
not buy any skirt.
Verify the answer by checking whether it satisfies the conditions of the given
problem. \
EXERCISE 3.1
1. Form the pair of linear equations in the following problems, and find their solutions
graphically.

(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4
more than the number of boys, find the number of boys and girls who took part in
the quiz.

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Pair OF LINEAR EQUATIONS IN TWo VARIABLES 29
(i) 5 pencils and 7 pens together cost = 50, whereas 7 pencils and 5 pens together
cost % 46. Find the cost of one pencil and that of one pen.
b
2. On comparing the ratios A, b and , find out whether the lines representing the
az By C2
following pairs of linear equations intersect at a point, are parallel or coincident:
@) 5x-4y+8=0 (i) 9x+3y+12=0
7x + 6y-9=0 18x + 6y + 24=0
(ii) 6x-3y+10=0 :
2x-y+9=0
. : a,b, CQ. : . : |
3. Oncomparing the ratios a bo and oe find out whether the following pair of linear
2 2 2 .
equations are consistent, or inconsistent. . . :
@) 3x+2y=5; 2x-3y=7 - di) 2x-3y=8;° 4x-6y=9
(iii) = + > =7;9x-l0y=14 > (iv) 5x-3y=11; -10x+ 6y=-22
4 . :
(v) att 2y =8 ; 2x4+3y=12 .
4, Which of the following pairs of linear equations are consistent/inconsistent? If
consistent, obtain the solution graphically:
@ xt+y=5, { .2e+2y=10 :
(i) x-y=8, "~~ 3x-3y=16
(ii) 2x+y-6=0, 4x-2y-4=0
(tv) 2x-2y-2=0, 4x-4y-5=0
5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is
36 m. Find the dimensions of the garden.
6. Given the linear equation 2x + 3y- 8 = 0, write another linear equation in two variables
such that the geometrical representation of the pair so formed is:
(i) intersecting lines (ii) parallel lines
(iit) coincident lines
7. Draw the graphs of the equations x - y + 1 = 0 and 3x + 2y - 12 = 0. Determine the
coordinates of the vertices of the triangle formed by these lines and the x-axis, and
shade the triangular region.
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30 MATHEMATICS
3.3 Algebraic Methods of Solving a Pair of Linear Equations
In the previous section, we discussed how to solve a pair of linear equations graphically.
The graphical method is not convenient in cases when the point representing the
solution of the linear equations has non-integral coordinates like (V3 ,2N7 ) ;
4 1 ; oa: ae ; .
(-1.75, 3.3), (4 5) , etc. There is every possibility of making mistakes while reading
such coordinates. Is there any alternative method of finding the solution? There are
several algebraic methods, which we shall now discuss.
3.3.1 Substitution Method : We shall explain the method of substitution by taking
some examples. -
Example 4: Solve the following pair of equations by substitution method:
7x --15y = 2 : oe (1)
x+2y=3 © : (2)
Solution : . _ .
Step 1 : We pick either of the equations and write one variable in terms of the other.
Let us consider the Equation (2) :
xt2y=30
and write it as ~~) x=3-2y (3)
Step 2 : Substitute the value of x in Equation (1). We get
7(3 -2y) -- 1Sy = 2
ie., 21- 14y- 15y = 2
ie, LO ~29y = -19
Theref 2
erefore, . Y= 59
Step 3 : Substituting this value of y in Equation (3), we get
- 19) 49
= 3-2)/-/=-
* ) 29
Therefore, the solution i * 2
erefore, the solution is x= 55. = 59:
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Pair OF LINEAR EQUATIONS IN TWo VARIABLES 31
tee gs mae 49 19 ; ,
Verification : Substituting x= 29 and y= 29° you can verify that both the Equations

(1) and (2) are satisfied.
To understand the substitution method more clearly, let us consider it stepwise:

Step 1 : Find the value of one variable, say y in terms of the other variable, i.c., x from
either equation, whichever is convenient.
Step 2: Substitute this value of y in the other equation, and reduce it to an equation in
one variable, i.e., in terms of x, which can be solved. Sometimes, as in Examples 9 and
10 below, you can get statements with no variable. If this statement is true, you can
conclude that the pair of linear equations has infinitely many solutions. If the statement
is false, then the pair of linear equations is inconsistent. . °
Step 3 : Substitute the value of x (or y) obtained in Step 2 in the equation used in
Step 1 to obtain the value of the other variable. oo
Remark : We have substituted the value of one variable by expressing it in terms of
the other variable to solve the pair of linear equations. That is why the method is
known as the substitution method. .
Example 5 : Solve the following question-Aftab tells his daughter, "Seven years
ago, I was seven times as old as you were then. Also, three years from now, I shall be
three times as old as you will be." (Isn't this interesting?) Represent this situation
algebraically and graphically by the method of substitution.
Solution : Let s and t be the ages (in years) of Aftab and his daughter, respectively.
Then, the pair of linear equations that represent the situation is

s-7=7(t-7), ie. s-7t+42=0 (1)
and s+3=3(t+3), ie, s-3r=6 (2)
Using Equation (2), we get s = 31+ 6.
Putting this value of s in Equation (1), we get

, (3+ 6)-7t+42= 0,
i.e., ° 4t= 48, which gives t= 12.
Putting this value of ¢ in Equation (2), we get
s=3(12)+6=42
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32 MATHEMATICS
So, Aftab and his daughter are 42 and 12 years old, respectively.

Verify this answer by checking if it satisfies the conditions of the given problems.
Example 6: Ina shop the cost of 2 pencils and 3 erasers is 79 and the cost of 4
pencils and 6 erasers is $18. Find the cost of each pencil and each eraser.

Solution : The pair of linear equations formed were:
2x+3y=9 (1)
4x + 6y = 18 _ @)
We first express the value of x in terms of y from the equation 2x + 3y = 9, to get
9-3y --
= -- - 63
x 7 (3)
Now we substitute this value of x in Equation (2), to get . . _
4(9 - 3y) ns
--- + b6y= 18 °
2 ne :
1.¢., 18 -6y + 6y = 18° _
1.¢., 18=18

This statement is true for all values of y. However, we do not get a specific
value of y as a solution. Therefore, we cannot obtain a specific value of x. This
situation has arisen because both the given equations are the same. Therefore,
Equations (1) and (2) have infinitely many solutions, We cannot find a unique
cost of a pencil and an eraser, because there are many common solutions, to the
given situation.

Example 7 : Two rails are represented by the equations
x+2y-4=0 and 2x+4y~12=0. Will the rails cross each other?
Solution : The pair of linear equations formed were:
x+2y-4=0 (1)
. 2x+4y-12=0 (2)
We express x in terms of y from Equation (1) to get
x=4-2y
Now, we substitute this value of x in Equation (2) to get
2(4 -2y) + 4y-12=0
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Pair OF LINEAR EQUATIONS IN TWo VARIABLES 33
Le, 8-12=0
Le, -4=0
which is a false statement.
Therefore, the equations do not have a common solution. So, the two rails will not
cross each other.
EXERCISE 3.2
1. Solve the following pair of linear equations by the substitution method. .
@ x+y=14 (i) s-t=3 ;
s ot
-~y= -+-=6
x-y=4 3°92 :
dil) 3x-y=3 (iv) 0.2x403y=13 > |
9x-3y=9 . 0.4x+ 0.5y = 2.3
. _ 3x -5y :
(v) V2. x+v3 y=0 ~ (vi) 703
x y 13
_ = =-42=-
V3x-8 y=0 3727S
2. Solve 2x + 3y = 11 and 2x - 4y = - 24 and hence find the value of 'm' for which
y=mrt+3. - _
3. Form the pair of linear equations for the following problems and find their solution by
substitution method. ,/ .
(i) The difference between two numbers is 26 and one number is three times the other.
Find them.
(i) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find
them. 7
(ui) The coach of a cricket team buys 7 bats and 6 balls for % 3800. Later, she buys 3
bats and 5 balls for 1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the
_ distance covered. For a distance of 10 km, the charge paid is ¥ 105 and for a
journey of 15 km, the charge paid is ¥ 155. What are the fixed charges and the
charge per km? How much does a person have to pay for travelling a distance of
25km? 9
(v) A fraction becomes li , if 2 is added to both the numerator and the denominator.
If, 3 is added to both the numerator and the denominator it becomes = . Find the
fraction.
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34 MATHEMATICS
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years
ago, Jacob's age was seven times that of his son. What are their present ages?
3.3.2 Elimination Method
Now let us consider another method of eliminating (1.e., removing) one variable. This
is Sometimes more convenient than the substitution method. Let us see how this method
works.
Example 8 : The ratio of incomes of two persons is 9 : 7 and the ratio of their
expenditures is 4 : 3. If each of them manages to save = 2000 per month, find their
monthly incomes. ;
Solution : Let us denote the incomes of the two person by % 9x and & 7x and their
expenditures by % 4y and % 3y respectively. Then the equations formed in the situation
is given by: ° . _
9x - 4y = 2000 | oe (1)
and Tx -3y= 2000 - : o (2)
Step 1: Multiply Equation (1) by 3 and Equation (2) by 4 to make the coefficients of
y equal. Then we get the equations: .
27x - 12y = 6000 © (3)
28x -12y = 8000 - (4)
Step 2: Subtract Equation (3) from Equation (4) to eliminate y, because the coefficients
of y are the same. So, we get
(28x - 27x) - (12y - 12y) = 8000-6000
1.¢., _ x= 2000
Step 3 : Substituting this value of x in (1), we get
* 9(2000) - 4y = 2000
i.e., SO y= 4000
So, the solution of the equations is x = 2000, y = 4000. Therefore, the monthly incomes
of the persons are = 18,000 and & 14,000, respectively.
Verification : 18000 : 14000 = 9: 7. Also, the ratio of their expenditures =
18000 - 2000 : 14000 - 2000 = 16000 : 12000 = 4:3
Remarks :
1. The method used in solving the example above is called the elimination method,
because we eliminate one variable first, to get a linear equation in one variable.
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Pair OF LINEAR EQUATIONS IN TWo VARIABLES 35
In the example above, we eliminated y. We could also have eliminated x. Try
doing it that way.

2. You could also have used the substitution, or graphical method, to solve this
problem. Try doing so, and see which method is more convenient.
Let us now note down these steps in the elimination method :

Step 1 : First multiply both the equations by some suitable non-zero constants to make

the coefficients of one variable (either x or y) numerically equal.

Step 2 : Then add or subtract one equation from the other so that one variable gets

eliminated. If you get an equation in one variable, go to Step 3. ,

If in Step 2, we obtain a true statement involving no variable, then the original

pair of equations has infinitely many solutions. a
If in Step 2, we obtain a false statement involving no variable, then the original

pair of equations has no solution, i.e., it is inconsistent. .

Step 3 : Solve the equation in one variable (x or y) so obtained to get its value.

Step 4 : Substitute this value of x (or y) in either of the original equations to get the

value of the other variable. oe "

Now to illustrate it, we shall solve few more examples.

Example 9 : Use elimination method to find all possible solutions of the following pair

of linear equations : . .

: 2x +3y = 8 (1)
Cc) dx+6y= 7 (2)

Solution : "

Step 1: Multiply Equation (1) by 2 and Equation (2) by | to make the

coefficients of x equal. Then we get the equations as :

. ~ 4x+b6by= 16 (3)
SO 4x + 6y=7 (4)
Step 2 : Subtracting Equation (4) from Equation (3),
_ (Ax - 4x) + (6y - 6y) = 16-7

Le., - 0 = 9, which is a false statement.

Therefore, the pair of equations has no solution.

Example 10 : The sum of a two-digit number and the number obtained by reversing

the digits is 66. If the digits of the number differ by 2, find the number. How many such

numbers are there?
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36 MATHEMATICS
Solution : Let the ten's and the unit's digits in the first number be x and y, respectively.
So, the first number may be written as 10x + y in the expanded form (for example,
56 = 10(5) + 6).

When the digits are reversed, x becomes the unit's digit and y becomes the ten's
digit. This number, in the expanded notation is 10y + x (for example, when 56 is
reversed, we get 65 = 10(6) + 5).

According to the given condition.
(10x + y) + (10y + x) = 66
Le., 11+ y) = 66 "
Le., x+y=6 (1)
We are also given that the digits differ by 2, therefore, os
either x-y=2 ; : (2)
or yrxs2 ", : (3)
If x - y = 2, then solving (1) and (2) by elimination, we get x =4and y=2.
In this case, we get the number 42. a oe "
If y -x = 2, then solving (1) and (3) by elimination, we get x=2 and y=4.
In this case, we get the number 24.
Thus, there are two such numbers 42 and 24.
Verification : Here 42 + 24 = 66 and 4 - 2 = 2. Also 24 + 42 = 66 and 4-2 =2.
o
NO EXERCISE 3.3
1. Solve the following pair of linear equations by the elimination method and the substitution
method :
@ x+y=5 and 2x-3y=4 (i) 3x+4y=10 and 2x-2y=2
. . x 2
(iii) 3x-Sy-4=0 and 9x=2y+7 (iv) stp a ctand a5 =3
2. Form the pair of linear equations in the following problems, and find their solutions
(if they exist) by the elimination method :
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces
1
to 1. It becomes 5 if we only add 1 to the denominator. What is the fraction?
(i) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as
old as Sonu. How old are Nuri and Sonu?
(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is
twice the number obtained by reversing the order of the digits. Find the number.
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Pair OF LINEAR EQUATIONS IN TWo VARIABLES 37
(tv) Meena went to a bank to withdraw % 2000. She asked the cashier to give her
= 50 and = 100 notes only. Meena got 25 notes in all. Find how many notes of

= 50 and & 100 she received.

(v) A lending library has a fixed charge for the first three days and an additional charge
for each day thereafter. Saritha paid ¥ 27 for a book kept for seven days, while Susy
paid % 21 for the book she kept for five days. Find the fixed charge and the charge
for each extra day.

3.4 Summary
In this chapter, you have studied the following points: :
1. A pair of linear equations in two variables can be represented, and solved, by the: -
() graphical method a
(ii) algebraic method .
2. Graphical Method : " , .
The graph of a pair of linear equations in two variables is represented by two lines.
(@) If the lines intersect at a point, then that point gives the unique solution of the two
equations. In this case, the pair of equations is consistent.

(i) Ifthe lines coincide, then there are infinitely many solutions - each point on
the line being a solution. In this case, the pair of equations is dependent
(consistent).

(iii) If the lines are parallel, then the pair of equations has no solution. In this case, the
pair of equations is inconsistent.

! s | .
3. Algebraic Methods : We have discussed the following methods for finding the solution(s)
of a pair of linear equations :
(i) Substitution Method
(ii) Elimination Method
4, Ifa pair of linear equations is given by a,x + b,y +c, =0 and a,x + b,y +c, = 0, then the
following situations can arise :
b . . . . :
(i) a4 b : In this case, the pair of linear equations is consistent.
ay 1
ee . : . . oe :
(ii) --=7 *- :In this case, the pair of linear equations is inconsistent.
a, db
see a, b, Cy . : : : . :
(ii) -= b =- :Inthis case, the pair of linear equations is dependent and consistent.
ag Op Cy
5. There are several situations which can be mathematically represented by two equations
that are not linear to start with. But we alter them so that they are reduced to a pair of
linear equations.
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fe) el

ey

icra

[a] ee

1062CHO4
4.1 Introduction .
In Chapter 2, you have studied different types of polynomials. One type was the
quadratic polynomial of the form ax? + bx + c,a #0. When we equate this polynomial
to zero, we get a quadratic equation. Quadratic equations come up when we deal with
many real-life situations. For instance, suppose a
charity trust decides to build a prayer hallhaving |,-2 ~7-- °° -- 0 -- |
a carpet area of 300 square metres with its length SP. 300m?
one metre more than twice its breadth. What 3. SO
should be the length and breadth of the hall? | |: 70 oO 7 7
Suppose the breadth of the hall is x metres. Then, x +1
its length should be (2x + 1) metres. We can depict .
this information pictorially as shown in Fig. 4.1. Fig. 4.1
Now, area of the hall = (2x + 1). xm' = (2? + x) m'
So, 2x7 +x = 300 (Given)
Therefore, 2x7? +x--300= 0
So, the breadth of the hall should satisfy the equation 2x? + x - 300 = 0 which is a
quadratic equation. ,
Many people believe that Babylonians were the first to solve quadratic equations.

For instance, they knew how to find two positive numbers with a given positive sum
and a given positive product, and this problem is equivalent to solving a quadratic
equation of the form x? - px + q = 0. Greek mathematician Euclid developed a
geometrical approach for finding out lengths which, in our present day terminology,
are solutions of quadratic equations. Solving of quadratic equations, in general form, is
often credited to ancient Indian mathematicians. In fact, Brahmagupta (C.E.598-665)
gave an explicit formula to solve a quadratic equation of the form ax' + bx = c. Later,

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QUADRATIC EQUATIONS 39
Sridharacharya (C.E. 1025) derived a formula, now known as the quadratic formula,
(as quoted by Bhaskara ID) for solving a quadratic equation by the method of completing
the square. An Arab mathematician Al-Khwarizmi (about C.E. 800) also studied
quadratic equations of different types. Abraham bar Hiyya Ha-Nasi, in his book

"Liber embadorum' published in Europe in C.E. 1145 gave complete solutions of

different quadratic equations.

In this chapter, you will study quadratic equations, and various ways of finding
their roots. You will also see some applications of quadratic equations in daily life
situations. .

4.2 Quadratic Equations

A quadratic equation in the variable x is an equation of the form ax'? + bx +c =0, where

a, b, c are real numbers, a # 0. For example, 2x* + x - 300 = 0 is a quadratic equation.

Similarly, 2x° - 3x + 1 = 0, 4x - 3x° + 2 = 0 and 1 - x° + 300 = 0 are also quadratic

equations, . --

In fact, any equation of the form p(x) = 0, where p(x) is a polynomial of degree
2, is a quadratic equation. But when we write the terms of p(x) in descending order of
their degrees, then we get the standard form of the equation. That is, ax? + bx +c=0,
a # (is called the standard form of a quadratic equation.

Quadratic equations arise in several situations in the world around us and in
different fields of mathematics. Let us consider a few examples.

a

Example 1 : Represent the following situations mathematically:

G) Johnand Jivanti together have 45 marbles. Both of them lost 5 marbles each, and
the product of the number of marbles they now have is 124. We would like to find
out how many marbles they had to start with.

(ii) A cottage industry produces a certain number of toys in a day. The cost of
production of each toy (in rupees) was found to be 55 minus the number of toys
produced in a day. On a particular day, the total cost of production was
= 750. We would like to find out the number of toys produced on that day.

Solution :

(i) Let the number of marbles John had be x.

Then the number of marbles Jivanti had = 45 - x (Why?).

The number of marbles left with John, when he lost 5 marbles = x -5

The number of marbles left with Jivanti, when she lost 5 marbles = 45 -- x -5

= 40-x
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40 MATHEMATICS
Therefore, their product = (x - 5) (40 - x)
= 40x - x' - 200 + 5x
= --x° + 45x - 200
So, -x° + 45x-200 = 124 (Given that product = 124)
ie, -2°+45x-324=0
1.¢., x°- 45x + 324= 0
Therefore, the number of marbles John had, satisfies the quadratic equation
45x + 324=0 :
which is the required representation of the problem mathematically.
(ti) Let the number of toys produced on that day be x. . :
Therefore, the cost of production (in rupees) of each toy that day =55-x
So, the total cost of production (in rupees) that day = x (55 - x)
Therefore, x(55-x)= 750 7
ie, 55x-2 = 750
ie. ~ x24 55x-750 = 0 po
1.¢., x -- 55x + 750 = 0
Therefore, the number of toys produced that day satisfies the quadratic equation
(= 55x+750= 0
which is the required representation of the problem mathematically.
Example 2 : Check whether the following are quadratic equations:
(i) @-2yP4+1=2x-3 (ii) xv t+ 1)+8=(4 + 2) @--2)
(iii) x Qx+3)=x7+1 (iv) («+ 2P=x-4
Solution : sO ;
@) LHS =@-2%4+1=x-4%4+44+1=%-4x4+5
Therefore, (x - 2)?+ 1 = 2x - 3 can be rewritten as
| 4x45 = 2x-3
i.e., x°>-6x+8=0
It is of the form ax? + bx +c =0.
Therefore, the given equation is a quadratic equation.
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QuapraTic EQUATIONS 41
Gi) Since x(x +1) +8 =2x?+x48 and (x + 2)(x-2) =x°-4

Therefore, vP+x4+8= 7-4

1.¢., x+12=0

It is not of the form ax? + bx +c =0.

Therefore, the given equation is not a quadratic equation.

(iii) Here, LHS = x (2x + 3) = 2x? + 3x
So, x (2x +3) = x? + 1 can be rewritten as
2° 4+ 3x= 41 '

Therefore, we get »°+3x-1=0 .

It is of the form ax? + bx +c =0. 7

So, the given equation is a quadratic equation. "

(iv) Here, LHS = (x + 27 = x7 + 6x7 + 12x +8 . _

Therefore, (x + 2) = x*- 4 can be rewritten as

O+62412x4+8=R-4- |. _

Le., 6x°+12x+12=0 or, x°4+2x4+2=0

It is of the form ax? + bx+c=0. . : :

So, the given equation is a quadratic equation.

Remark : Be careful! In (ii) above, the given equation appears to be a quadratic
equation, but it is not a quadratic equation.

In (iv) above, the given equation appears to be a cubic equation (an equation of
degree 3) and not a quadratic equation. But it turns out to be a quadratic equation. As
you can see, often we need to simplify the given equation before deciding whether it
is quadratic or not.

~ | EXERCISE 4.1
1. Check whether the following are quadratic equations :
@) (x+1)?=2(¢-3) (ii) x? -2x = (-2) (3-x)
(mi) (x¥-2)@4+ D=@-D@+3) (tv) (x-3)(2x4+1) =x(x +5)
(vy) Qx-Dx-3) = 5)(x-D (vi) x2+3x+1=(x-2)
(vil) (x + 2) = 2x (x?- 1) (viii) 0° -- 4? -x + 1 =(x-2)
2. Represent the following situations in the form of quadratic equations :
(i) The area of a rectangular plot is 528 m'. The length of the plot (in metres) is one
more than twice its breadth. We need to find the length and breadth of the plot.
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42 MATHEMATICS
(i) The product of two consecutive positive integers is 306. We need to find the
integers.
(iii) Rohan's mother is 26 years older than him. The product of their ages (in years)
3 years from now will be 360. We would like to find Rohan's present age.
(tv) A train travels a distance of 480 km at a uniform speed. If the speed had been
8 km/h less, then it would have taken 3 hours more to cover the same distance. We
need to find the speed of the train.
4.3 Solution of a Quadratic Equation by Factorisation
Consider the quadratic equation 2x? - 3x + 1 = 0. If we replace x by 1 on the
LHS of this equation, we get (2 x 1') -- (3 x 1) + 1 = 0 = RHS of the equation.
We say that 1 is a root of the quadratic equation 2x? - 3x + 1 =0. This also means that
1 is a zero of the quadratic polynomial 2x? - 3x + 1. .

In general, a real number © is called a root of the quadratic equation
ax + bx+c=0,a40if ao? + ba+c=0. We also say that x = & is a solution of
the quadratic equation, or that o satisfies the quadratic equation. Note that the
zeroes of the quadratic polynomial ax' + bx + ¢ and the roots of the quadratic
equation ax? + bx +c =0 are the same. .~ °

You have observed, in Chapter 2, that a quadratic polynomial can have at most
two zeroes. So, any quadratic equation can have atmost two roots.

You have learnt in Class IX, how to factorise quadratic polynomials by splitting
their middle terms. We shall use this knowledge for finding the roots of a quadratic
equation. Let us see how. ) °

ML .
Example 3 : Find the roots of the equation 2x* - 5x + 3 = 0, by factorisation.
Solution : Let us first split the middle term - 5x as --2x -3x [because (-2x) x (-3x) =
6x? = (2x°) x 3).
So, 2x? - 5x +3 = 2x?-2x - 3x 43 = 2x (x- 1) -3(e- 1) = (2x - 3)(x- 1)
Now, 2x? - 5x + 3 = 0 can be rewritten as (2x - 3)(x - 1) =0.
So, the values of x for which 2x2 - 5x + 3 =0 are the same for which (2x -3)(x- 1) =0,
1.e., either 2x -3 =Oorx-1=0.
: 3
Now, 2x - 3 = 0 gives xey and x- 1 =0 gives x= 1.
3
So, ¥= 2 and x = | are the solutions of the equation.
3
In other words, 1 and 3 are the roots of the equation 2x? -5x+3=0.
Verify that these are the roots of the given equation.
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QuADRATIC EQUATIONS 43
Note that we have found the roots of 2x? - 5x + 3 = 0 by factorising
2x° - 5x + 3 into two linear factors and equating each factor to zero.
Example 4 : Find the roots of the quadratic equation 6x* - x -2 = 0.
Solution : We have
6x? -x--2=6x° 4+ 3x-4x-2

= 3x (2x + 1)-2 (2x + 1)

= (3x --2)(2x + 1)
The roots of 6x* - x - 2 = 0 are the values of x for which (3x - 2)(2x + 1) =0
Therefore, 3x - 2 = 0 or 2x + 1 =0, .
; 2 I
Le., x= 3 or x= ZO - 7

2 0 4
Therefore, the roots of 6x° - x - 2 = 0 are 3 and 5" _
| 2d
We verify the roots, by checking that 3 and "5 satisfy 6x? -x-2=0.
Example 5 : Find the roots of the quadratic equation 3x7 - 2/6 y+2=0.
Solution : 3x? -2/6x+ oy 3x? -J6x-J6x+2
ML .
= V3 x(V3 x - V2) ~ 2(V3x - V2}
= (V3x~V2)(J3x-2)
So, the roots of the equation are the values of x for which
~ (V3x-V2)(V3x- v2) =0
2
Now, /3x-J2 =0 for v=.
So, this root is repeated twice, one for each repeated factor V3x-V2.
2 j|2
Therefore, the roots of 37 - w6 x+2=0 are 3° V3 :
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44 MATHEMATICS
Example 6 : Find the dimensions of the prayer hall discussed in Section 4.1.
Solution : In Section 4.1, we found that if the breadth of the hall is x m, then x
satisfies the equation 2x? + x - 300 = 0. Applying the factorisation method, we write
this equation as
2x° - 24x + 25x - 300 = 0
2x (x -- 12) + 25 @-12)=0
Le., (x - 12)(2x +25) = 0
So, the roots of the given equation are x = 12 or x =- 12.5. Since xis the breadth
of the hall, it cannot be negative.
Thus, the breadth of the hall is 12 m. Its length = 2x + 1 = 25 m. .
EXERCISE 4.2 oO
1. Find the roots of the following quadratic equations by factorisation: .
@ x°-3x-10=0 Gi) W+x-6=0 *
Lo 1
Git) J2.x?+7x+5V2=0 " (iv) 2eP-x4 2 =0
(v) 100x?-20x+1=0 . ,
2. Solve the problems given in Example 1.
3. Find two numbers whose sum is 27 and product is 182.
4. Find two consecutive positive integers, sum of whose squares is 365.
5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find
the other two sides. °
6. Acottage industry produces a certain number of pottery articles in a day. It was observed
ona particular day that the cost of production of each article (in rupees) was 3 more than
twice the number of articles produced on that day. If the total cost of production on that
day was % 90, find the number of articles produced and the cost of each article.
4.4 Nature of Roots
The equation ax' + bx + c = 0 are given by
- b+, Jb? -4ac
; x= -
2a
oe b Vb' -4ac
If b? - 4dac > 0, we get two distinct real roots Sa + a and
a a
b \b' -4ac
2a 2a
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QUADRATIC EQUATIONS 45

If b? - 4ac = 0, then x = -? 40, ic, x=-2 or 2.

CEM Men X= 2a 2a 2a
. -b
So, the roots of the equation ax? + bx + c = 0 are both Da
a

Therefore, we say that the quadratic equation ax' + bx + c = 0 has two equal
real roots in this case.

If b? -4ac < 0, then there is no real number whose square is b? - 4ac. Therefore,
there are no real roots for the given quadratic equation in this case.

Since b? - 4ac determines whether the quadratic equation ax? + bx +c = 0 has
real roots or not, b* - 4ac is called the discriminant of this quadratic equation.
So, a quadratic equation ax? + bx+c=Ohas_. :

(i) two distinct real roots, if b? - 4ac > 0, oe
(ii) two equal real roots, if b* -- 4dac = 0, _ -
(iii) no real roots, if b?-4ac <0. =. / :

Let us consider some examples. . ° ,
Example 7: Find the discriminant of the quadratic equation 2x? - 4x + 3 = 0, and
hence find the nature of its roots. .
Solution : The given equation is of the form ax? + bx +c = 0, where a = 2, b=-4 and
c =3. Therefore, the discriminant °

SY
b? - 4ac = (- 4)' - (4x 2 x 3) = 16-24=-8<0

So, the given equation has no real roots.
Example 8 : A pole has to be erected at a point on the boundary of a circular park of
diameter 13 metres in such a way that the differences of its distances from two
diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to
do so? If yes, at what distances from the two gates should the pole be erected?
Solution : Let us first draw the diagram Fy
(see Fig. 4.2). . { [ell 'BY

Let P be the required location of the a} Ito
pole. Let the distance of the pole from the (3 ;
gate B be x m, i.e., BP = x m. Now the 'al [I~
difference of the distances of the pole from
the two gates = AP - BP (or, BP - AP) = P
7m. Therefore, AP = (x + 7) m. is

Fig. 4.2
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46 MATHEMATICS

Now, AB = 13m, and since AB is a diameter,

ZAPB = 90° (Why?)
Therefore, AP? + PB? = AB? (By Pythagoras theorem)
1.¢., (x+7P4+x°= 13?
ie., x + 14% +49 +x? = 169
1.¢., 2x° + 14x-120= 0
So, the distance 'x' of the pole from gate B satisfies the equation
x + 7x-60= 0 °
So, it would be possible to place the pole if this equation has real roots. To see if this
is so or not, let us consider its discriminant. The discriminant is
b? -4ac = 7T?- 4x 1 x (- 60) = 289 > 0. .

So, the given quadratic equation has two real roots, and it is possible to erect the
pole on the boundary of the park. , Oe

Solving the quadratic equation x + 7x - 60 =0, by the quadratic formula, we get

~ 74/289 -7 £17
x Se oT =" DD ©
2 2

Therefore, x = 5 or - 12.

Since x is the distance between the pole and the gate B, it must be positive.
Therefore, x = - 12 will have to be ignored. So, x= 5.

4

Thus, the pole has to be erected on the boundary of the park at a distance of 5m
from the gate B and 12m from the gate A.
Example 9 : Find the discriminant of the equation 3x° - 2x + 3 0 and hence find the
nature of its roots. Find them, if they are real.

. ; 1
Solution : Here a = 3, b = - 2 and aa'
4, 1
Therefore, discriminant b? -- 4ac = (-2)?-4x3 x 3 = 4-4=0.
Hence, the given quadratic equation has two equal real roots.
€ roots are FB Ma er ee Meo 3g
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QUADRATIC EQUATIONS 47
EXERCISE 4.3

1. Find the nature of the roots of the following quadratic equations. If the real roots exist,
find them:

@) 2x°-3x4+5=0 (ii) 3x°-4 V3 x+4=0
(iii) 2x?-6x+3=0

2. Find the values of k for each of the following quadratic equations, so that they have two
equal roots.

(i) 2x? +kx+3=0 (i) kx (x-2)+6=0 °

3. Is it possible to design a rectangular mango grove whose length is twice its breadth,
and the area is 800 m?? If so, find its length and breadth.

4. Is the following situation possible? If so, determine their present ages.

The sum of the ages of two friends is 20 years. Four years ago, the product of their ages
in years was 48. . oe

5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m?? If so, find
its length and breadth. a , ° "

4.5 Summary °
In this chapter, you have studied the following points:

1. A quadratic equation in the variable x is of the form ax?+ bx + c=0, where a, b, c are real
numbers and a # 0. ~ .

2. A real number © is said to be a root of the quadratic equation ax? + bx + c = 0, if
ao? + ba +c =0. The zeroes of the quadratic polynomial ax? + bx + ¢ and the roots of the
quadratic equation ax? + bx + c= are the same.

3. If we can factorise ax? + bx +c, a#0, into a product of two linear factors, then the roots
of the quadratic equation ax' + bx + c=0 can be found by equating each factor to zero.

4. Quadratic formula: The roots of a quadratic equation ax? + bx + c = 0 are given by

-bt+/p? -
-b t vb" -4ac + vb" -4ac > provided b? - 4ac = 0.
2a
5. A quadratic equation ax? + bx +c =0 has

(i) two distinct real roots, if b?-4ac > 0,

(ii) two equal roots (i.e., coincident roots), if b? - 4ac = 0, and
(iii) no real roots, if b?--4ac <0.

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aed
Ela
wha
a - wu
al
(a) ter
1062CHO5
5.1 Introduction .
You must have observed that in nature, many things follow a certain pattern, such as
the petals of a sunflower, the holes of a honeycomb, the grains on a maize cob, the
spirals on a pineapple and on a pine cone, etc. . . _
We now look for some patterns which occur in our day-to-day life. Some such
examples are : °
(i) Reena applied for a job and got selected. She
has been offered a job with a starting monthly
salary of = 8000, with an annual increment of
= 500 in her salary. Her salary (in %) for the Ist,
2nd, 3rd, . . . years will be, respectively
8000, 8500,- 9000,....
(ii) The lengths of the rungs of a ladder decrease
uniformly by 2 cm from bottom to top
(see Fig. 5.1). The bottom rung is 45 cm in
length. The lengths Gin cm) of the Ist, 2nd,
3rd, .. ., 8th rung from the bottom to the top
are, respectively Fig. 5.1
45, 43, 41, 39, 37, 35, 33, 31
(ii) In a savings scheme, the amount becomes 4 times of itself after every 3 years.
The maturity amount (in %) of an investment of % 8000 after 3, 6, 9 and 12 years
will be, respectively :
10000, 12500, 15625, 19531.25
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50 MATHEMATICS

(iv) The number of unit squares in squares with side 1, 2, 3,... units (see Fig. 5.2)
are, respectively

17,2°,37,....
Fig. 5.2 / :

(v) Shakila puts = 100 into her daughter's money box when she was one year old
and increased the amount by % 50 every year. The amounts of money (in %) in the
box on the Ist, 2nd, 3rd, 4th,... birthday were _

100, 150, 200, 250,..., respectively, °

(vi) A pair of rabbits are too young to produce in their first month. In the second, and
every subsequent month, they produce a new pair. Each new pair of rabbits
produce a new pair in their second month and in every subsequent month (see
Fig. 5.3). Assuming no rabbit dies, the number of pairs of rabbits at the start of
the Ist, 2nd, 3rd, ..., 6th month, respectively are :

\ S .
1,1, 2,3,5,8
1 |
~
2
3

7 oN, 7 ¥ oN,

+ Ne , + ™ +s v
8

Fig. 5.3
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ARITHMETIC PROGRESSIONS 51

In the examples above, we observe some patterns. In some, we find that the
succeeding terms are obtained by adding a fixed number, in other by multiplying
with a fixed number, in another we find that they are squares of consecutive
numbers, and so on.

In this chapter, we shall discuss one of these patterns in which succeeding terms
are obtained by adding a fixed number to the preceding terms. We shall also see how
to find their nth terms and the sum of n consecutive terms, and use this knowledge in
solving some daily life problems.

5.2 Arithmetic Progressions '
Consider the following lists of numbers :
(i) 1, 2, 3, 4,... .
Gi) 100, 70, 40, 10,... . :
(ii) -3, -2, -1, 0,... . 7, .
(iv) 3, 3, 3, 3,... ; oe -
(v) -1.0, -1.5, -2.0, -2.5,... oO / .

Each of the numbers in the list is called a term.

Given a term, can you write the next term in each of the lists above? If so, how
will you write it? Perhaps by following a pattern or rule. Let us observe and write the
rule. _ - _

In (i), each term is 1 more than the term preceding it.

In (ii), each term is 30 less than the term preceding it.

In (iii), each term is obtained by adding 1 to the term preceding it.

In (iv), all the terms in the list are 3 , i.e., each term is obtained by adding
(or subtracting) 0 to the term preceding it.

In (v), each term is obtained by adding - 0.5 to (i.e., subtracting 0.5 from) the
term preceding it. ~

In all the lists above, we see that successive terms are obtained by adding a fixed
number to the preceding terms. Such list of numbers is said to form an Arithmetic
Progression ( AP ).

So, an arithmetic progression is a list of numbers in which each term is
obtained by adding a fixed number to the preceding term except the first
term.

This fixed number is called the common difference of the AP. Remember that
it can be positive, negative or zero.

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52 MATHEMATICS

Let us denote the first term of an AP by a,, second term by a,, . . ., nth term by
a, and the common difference by d. Then the AP becomes a,, a,, a,,..., 4,

So, a,- 4a, =d,-d,=...=a4,-a ,=d.
Some more examples of AP are:
(a) The heights (in cm ) of some students of a school standing in a queue in the

morning assembly are 147 , 148, 149, ..., 157.

(b) The minimum temperatures ( in degree celsius ) recorded for a week in the

month of January in a city, arranged in ascending order are .

- 3.1, -3.0, - 2.9, - 2.8, - 2.7, - 2.6, - 2.5
(c) The balance money (in @ ) after paying 5 % of the total loan of = 1000 every
month is 950, 900, 850, 800, .. ., 50. . :
(d) The cash prizes (in @ ) given by a school to the toppers of Classes I to XII are,

respectively, 200, 250, 300, 350, ..., 750. -- .

(ce) The total savings (in %) after every month for 10 months when @ 50 are saved

each month are 50, 100, 150, 200, 250, 300, 350, 400, 450, 500.

It is left as an exercise for you to explain why each of the lists above is an AP.

You can see that ;

a,at+d,at+2d,a+3d,...
represents an arithmetic progression where a is the first term and d the common
difference. This is called the general form of an AP.
)

Note that in examples (a) to (e) above, there are only a finite number of terms.
Such an AP is called a finite AP. Also note that each of these Arithmetic Progressions
(APs) has a last term. The APs in examples (i) to (v) in this section, are not finite APs
and so they are called infinite Arithmetic Progressions. Such APs do not have a
last term. 7

Now, to know about an AP, what is the minimum information that you need? Is it
enough to know the first term? Or, is it enough to know only the common difference?
You will find that you will need to Know both - the first term a and the common
difference d. _

For instance if the first term a is 6 and the common difference d is 3, then
the AP is

6,9,12,15,...
and if a is 6 and d is - 3, then the AP is
6, 3, 0,-3,...
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ARITHMETIC PROGRESSIONS 53

Similarly, when

a=-7, d=-2, theAPis -7,-9,-11,-13,...
a= 1.0, d=0.1, the APis 1.0, 1.1, 1.2,1.3,...

0 d=1- he APis 0, 1~,3,4~,6
a= 0, =1,, the is 0, 15,3,45,6,...
a=2, d=0, the APis 2,2,2,2,...

So, if you know what a and d are, you can list the AP. What about the other way
round? That is, if you are given a list of numbers can you say that it is an AP and then
find a and d? Since a is the first term, it can easily be written. We know that in an AP,
every succeeding term is obtained by adding d to the preceding term. So, d found by
subtracting any term from its succeeding term, i.e., the term which immediately follows
it should be same for an AP. . .

For example, for the list of numbers: ~ a

6,9,12,15,..., - | 2
We have a,-a,=9 -6=3, -

a,-a,=12-9=3, ©

a,-a,=15--12=3

Here the difference of any two consecutive terms in each case is 3. So, the
given list is an AP whose first term a is 6 and common difference d is 3.

fry

For the list of numbers : 6, 3, 0,-3,.. 5

" g-a,=3~6=-3
a,-a,=0-3=-3
a,-a,=-3-0 =-3

Similarly this is also an AP whose first term is 6 and the common difference
is -3. ao ,

In general, for an AP a,,a,,...,@,, we have

d= 4, ,,- 4,
where a,,, and a, are the ( k + 1)th and the kth terms respectively.

To obtain d in a given AP, we need not find all of a,-a,,a,-a,,a,-d,,....
It is enough to find only one of them.

Consider the list of numbers 1, 1, 2, 3,5,.... By looking at it, you can tell that the
difference between any two consecutive terms is not the same. So, this is not an AP.

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54 MATHEMATICS
Note that to find d in the AP: 6, 3, 0,--3,..., we have subtracted 6 from 3
and not 3 from 6, i.e., we should subtract the kth term from the (k + 1) th term
even if the (k + 1) th term is smaller.
Let us make the concept more clear through some examples.
Example 1 : For the AP : 3. 1 - 1 _3 ,-.., Write the first term a and the
common difference d. a? ° °
Solution : Here a=2 doe 2-1.
, 2' 2 2
Remember that we can find d using any two consecutive terms, once we know that
the numbers are in AP. °
Example 2 : Which of the following list of numbers form an AP? If they form an AP,
write the next two terms : - , °
(i) 4, 10, 16, 22,... Gi) 1,-1,-3,-5,...0-
(iit) --2,2,-2,2,-2,... (iv) 1, 11, 2, 2, 2,3, 3,3,...
Solution: (i) We have a,-a, =10-4= 6 |
a,-a, =16-10= 6
mts 222-16 =6
Le., a,,,-4, Is the same every time. °
So, the given list of numbers forms an AP with the common difference d = 6.
The next two terms are: 22 + 6 = 28 and 28 + 6 = 34.
Gi) a,-a,=-1-1=-2 |
a,-a,= -3--(-1)=-3+1=-2
a,-a, =-5-(-3)=-54+3=-2
Le, a ,,-% is the same every time.
So, the given list of numbers forms an AP with the common difference d = --2.
The next two terms are:
--54(2)=-7 and -74+(02)=-9
(ill) a, -a,=2-(-2)=24+2=4
ad,-a,=-2-2=-4
Asa,-4d, # a,-a,, the given list of numbers does not form an AP.
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ARITHMETIC PROGRESSIONS 55
(iv) a,-a,=1-1=0
a,-a,=1-1=0
a,-a,=2-1=1
Here, a, - a, = a,- a, #4, -4,.
So, the given list of numbers does not form an AP.
EXERCISE 5.1
1. In which of the following situations, does the list of numbers involved make an arithmetic
progression, and why? °
(i) The taxi fare after each km when the fare is % 15 for the first km and ¥ 8 for each
additional km.
1
(i) The amount of air present in a cylinder when a vacuum pump removes - of the
air remaining in the cylinder atatime. - . _ 4
(iii) The cost of digging a well after every metre of digging, when it costs = 150 for the
first metre and rises by = 50 for each subsequent metre. =~
(iv) The amount of money in the account every year, when = 10000 is deposited at
compound interest at 8 % per annum. °
2. Write first four terms of the AP, when the first term a and the common difference d are
given as follows:
(i) a=10, d=10 . Gi) a=-2, d=0
- 1
(iii) a=4, ,4=-3 wy) a=-1, d= >
(v) a=- 1.25, | d=~ 0.25
3. For the following APs, write the first term and the common difference:
@) 3,1,-1,-3,... Gi) -5,-1,3,7,...
iy) - 2,2, 3 iv) 0.6, 1.7, 2.8, 3.9
M33 3.380 (iv) 0.6, 1.7, 2.8, 3.9,...
4, Which of the following are APs ? If they form an AP, find the common difference d and
write three more terms.
@) 2,4,8,16,... @) 2,237...
- 2 2
(wi) -1.2,-3.2,-5.2,-7.2,... (iv) -10,-6,-2, 2,...
(v) 3,3+V2,34+2V2,343V2,... (vi) 0.2,0.22, 0.222, 0.2222,...
i) O.-4,-8.-1 tht
(vii) 9 Fs Oy pees (il) - Bom BT eT
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56 MATHEMATICS
(x) 1,3,9,27,... (x) a, 2a, 3a, 4a,...
(xi) a,a',a?,a',... (xii) V2, V8, VI8, ¥32,.--
(xiii) V3, 6, JO, V2, ... (xiv) 17,37, 52,7...
(xv) 17,57, 77, 73,...
5.3 nth Term of an AP
Let us consider the situation again, given in Section 5.1 in which Reena applied for a
job and got selected. She has been offered the job with a starting monthly salary of
= 8000, with an annual increment of = 500. What would be her monthly salary for the
fifth year? .

To answer this, let us first see what her monthly salary for the second year
would be. .

It would be % (8000 + 500) = % 8500. In the same way, we can find the monthly
salary for the 3rd, 4th and 5th year by adding % 500 to the salary of the previous year.
So, the salary for the 3rd year = % (8500 +500) © - _

= = (8000 + 500 + 500) -
= %(8000+2x500)
=z [8000 + (3 - 1) x 500] (for the 3rd year)
==79000 .
Salary for the 4th year = Z (9000 + 500)
\ ) = = (8000 + 500 + 500 + 500)
= % (8000 + 3 x 500)
= = (8000 + (4 - 1) x 500] (for the 4th year)
. = 79500
Salary for the Sth year =z (9500 + 500)
_ ; = % (8000+500+500+500 + 500)
= = (8000 + 4 x 500)
" = % [8000 + (5 - 1) x 500] (for the 5th year)
" = % 10000
Observe that we are getting a list of numbers
8000, 8500, 9000, 9500, 10000, . . .
These numbers are in AP. (Why?)
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ARITHMETIC PROGRESSIONS 57
Now, looking at the pattern formed above, can you find her monthly salary for
the 6th year? The 15th year? And, assuming that she will still be working in the job,
what about the monthly salary for the 25th year? You would calculate this by adding
= 500 each time to the salary of the previous year to give the answer. Can we make
this process shorter? Let us see. You may have already got some idea from the way
we have obtained the salaries above.
Salary for the 15th year
= Salary for the 14th year + = 500 "
500 + 500 +500 + ...+500
= «| s000 + ----_- : = 500
13 times
= % [8000 + 14 x 500] . :
= = [8000 + (15-1) x 500) =% 15000. .
Le., First salary + (15 - 1) x Annual increment. ~
In the same way, her monthly salary for the 25th year would be
= (8000 + (25 - 1) x 500] = % 20000
= First salary + (25 - 1) x Annual increment
This example would have given you some idea about how to write the 15th term,
or the 25th term, and more generally, the nth term of the AP.
Let a,, a,, a,,... be an AP whose first term a, is a and the common
difference is d.
Then, -
the second term a, = a+d=a+(2-I1)d
thethirdterm a, = a, +d=(at+d)+d=a+2d=a+(3-1)d
the fourthterm a,= a,+d=(a+2d)+d=a+3d=a+(4-I)d
Looking at the pattern, we can say that the nth term a, =a + (n--1)d.
So, the nth term a, of the AP with first term a and common difference d is
given by a, =a + (n -1)d.
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58 MATHEMATICS

a, is also called the general term of the AP. If there are m terms in the AP, then
a,, tepresents the last term which is sometimes also denoted by /.

Let us consider some examples.
Example 3 : Find the 10th term of the AP: 2,7, 12,...
Solution : Here,@=2, d=7-2=5 and n= 10.
We have a,=at(n-Il1)d
So, a,=2+(0-1)x5=2+45=47
Therefore, the 10th term of the given AP is 47.
Example 4 : Which term of the AP: 21, 18, 15,...is -81? Also, is any term 0? Give
reason for your answer. -
Solution : Here, a= 21, d= 18-21 =-3 and a, = - 81, and we have to find n.
As a,=a+(n-1)d, po
wehave -81=21+(n-1)-3) > _

~81=24-3n ss
-~105=-3n : | |
So, n= 35
Therefore, the 35th term of the given AP is- 81.
Next, we want to know if there is any n for which a, = 0. If such an v7 is there, then
214 (n- 1) 3) = 0,
1.¢., 3(n - 1) = 21
Le., n= 8
So, the eighth term is 0. .
Example 5 : Determine the AP whose 3rd term is 5 and the 7th term is 9.
Solution : We have
° a,=at+@GB-I)d=a+2d=5 (1)
and : a,=at+(7-l)d=a+6d=9 (2)
Solving the pair of linear equations (1) and (2), we get
a= 3, d=1
Hence, the required AP is 3, 4,5,6,7,...
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ARITHMETIC PROGRESSIONS 59
Example 6 : Check whether 301 is a term of the list of numbers 5, 11, 17, 23,...
Solution : We have :

a,-a,=11-5=6, a,-a,=17-11=6, a,-a,=23-17=6
As a, ,,- 4, is the same for k = 1, 2, 3, etc., the given list of numbers is an AP.
Now, a=5 and d=6.
Let 301 be a term, say, the mth term of this AP.
We know that

a,=a+(n-l1)d .
So, 301=54+@-1)x6
ie. 301= 6n-1 | 4
302 151 .
So, n= 6 3 a |
But n should be a positive integer (Why?). So, 301 is not a term of the given list of
numbers. oe -
Example 7 : How many two-digit numbers are divisible by 3?
Solution : The list of two-digit numbers divisible by 3 is:
12, 15, 18,..., 99
Is this an AP? Yes it is. Here, a=12, d=3, a,=99.
As NA a, =at(n--l)d,
we have 99s 12+ (2-1) x3
i.e., 87 = (n-1)x3
i.e. gets Lao
, . ; 3
1.¢., n= 29+1=30
So, there are 30 two-digit numbers divisible by 3.
Example 8 : Find the 11th term from the last term (towards the first term) of the
AP: 10,7, 4,...,-62.
Solution : Here, a= 10, d=7-10=-3, /=- 62,
where l=a+(n-l)d
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60 MATHEMATICS

To find the | 1th term from the last term, we will find the total number of terms in
the AP.

So, -62=10+(@- 1)(-3)
Le., -72 = (n - 1)(-3)

Le., n--1=24

or n= 25

So, there are 25 terms in the given AP.

The 11th term from the last term will be the 15th term. (Note that it will not be
the 14th term. Why?)
So, a,, = 10+ (15 - 1)(-3) = 10-42 =- 32
i.e., the 11th term from the last term is - 32, "

Alternative Solution : : . . _

If we write the given AP in the reverse order, then a= _ 62 and d = 3 (Why?)

So, the question now becomes finding the 11th term with these a and d.
So, a,, = -62+ (11-1) x3=-62+30=-32
So, the 11th term, which is now the required term, is - 32. .
Example 9: A sum of = 1000 is invested at 8% simple interest per year. Calculate the
interest at the end of each year. Do these interests form an AP? If so, find the interest
at the end of 30 years making use of this fact.
Solution : We know that the formula to calculate simple interest is given by
PxRxT

Simple Interest = 100
So, the interest at the end of the Ist year = & "a = = 80
The interest at the end of the 2nd year= 2 Se =% 160
The interest at the end of the 3rd year= % - = = 240
Similarly, we can obtain the interest at the end of the 4th year, 5th year, and so on.
So, the interest (in %) at the end of the Ist, 2nd, 3rd, .. . years, respectively are

80, 160, 240,...
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ARITHMETIC PROGRESSIONS 61
It is an AP as the difference between the consecutive terms in the list is 80, i.e.,
d= 80. Also, a = 80.
So, to find the interest at the end of 30 years, we shall find a,,.
Now, a,, = a+ (30-1) d=80 + 29 x 80 = 2400
So, the interest at the end of 30 years will be = 2400.
Example 10 : In a flower bed, there are 23 rose plants in the first row, 21 in the
second, 19 in the third, and so on. There are 5 rose plants in the last row. How many
rows are there in the flower bed?
Solution ; The number of rose plants in the 1st, 2nd, 3rd, .. ., rows are : .
23,21, 19,...,5 / ,
It forms an AP (Why?). Let the number of rows in the flower bed be n. -
Then a=23, d=21-23=-2,4,55 0 9 |
As, a,=at(n-1)d oo -
We have, 5= 23 +(n-1)(-2) °
ie., ~18 = (n-1)(-2)- oe
i.e., n= 10 :
So, there are 10 rows in the flower bed. | © ,
(- + .
Ne EXERCISE 5.2
1, Fill in the blanks in the following table, given that a is the first term, d the common
difference and a, the nth term of the AP:
rr ee ee
(i) 7 3 8 vee
(ii) -18 vee 10 0
Gi) |... 3 18 5
(iv) | -18.9 2.5 vee 3.6
(v) 3.5 0 105 vee
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62 MATHEMATICS
2. Choose the correct choice in the following and justify :
(i) 30th term of the AP: 10, 7,4,..., is
(A) 97 (B) 77 (C) -77 (D) -87
(i) 11th term of the AP: - 3, 5 »2,+++,18
1
(A) 28 (B) 22 (C) -38 (D) -48 5
3. In the following APs, find the missing terms in the boxes :
02 [ |, 2
[| [| 3 |
] .
o OO |
(iii) > | SO
wa OOO. |
2 A
4. Which term of the AP:3,8,13,18,...,is 78?
5. Find the number of terms in each of the following APs :
s ' 1]
(i) 7,13, 19,...,205 . - (i) 18, 15-,13,...,-47
rN |
6. Check whether - 150 is aterm of the AP: 11, 8,5,2...
7, Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
8. AnAP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th
term. ,
9. Ifthe 3rd and the 9th terms of an AP are 4 and - 8 respectively, which term of this AP is
zero? , >
10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
11, Which term of the AP: 3, 15, 27, 39, ... will be 132 more than its 54th term?
12. Two APs have the same common difference. The difference between their 100th terms is
100, what is the difference between their 1000th terms?
13. How many three-digit numbers are divisible by 7?
14. How many multiples of 4 lie between 10 and 250?
15. For what value of n, are the nth terms of two APs: 63, 65, 67, ...and3, 10, 17,... equal?
16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
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ARITHMETIC PROGRESSIONS 63
17. Find the 20th term from the last term of the AP: 3, 8, 13,..., 253.
18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is
44. Find the first three terms of the AP.
19. Subba Rao started work in 1995 at an annual salary of € 5000 and received an increment
of % 200 each year. In which year did his income reach = 7000?
20. Ramkali saved & 5 in the first week of a year and then increased her weekly savings by
= 1.75. If in the nth week, her weekly savings become 20.75, find n.
5.4 Sum of First n Terms of an AP
Let us consider the situation again
given in Section 5.1 in which Shakila , ed N
. : V7 VV N .
put = 100 into her daughter's money TWAIIV-V aa .
box when she was one year old, Pm ip :
~ 150 on her second birthday, ye SO "
% 200 on her third birthday and will VEING eS
continue in the same way. How much A KX) =p 7
money will be collected in the money ' 4 (XN.
box by the time her daughter is 21 . . Uf: J?
years old? . "es

Here, the amount of money (in ) put in the money box on her first, second, third,
fourth .. . birthday were respectively 100, 150, 200, 250, ... till her 21st birthday. To
find the total amount in the money box on her 21st birthday, we will have to write each
of the 21 numbers in the list above and then add them up. Don't you think it would be
a tedious and time consuming process? Can we make the process shorter? This would
be possible if we can find a method for getting this sum. Let us see.

We consider the problem given to Gauss (about whom you read in
Chapter 1), to solve when he was just 10 years old. He was asked to find the sum of
the positive integers from 1 to 100. He immediately replied that the sum is 5050. Can
you guess how did he do? He wrote :

S=14+24+3+...+99+4 100
And then, reversed the numbers to write
- $= 1004+994+...434+24+1
Adding these two, he got
28S = (100+ 1) + (994+ 2)+...4+(3 + 98) 4+ (24+ 99) + (1 + 100)
= 101+101+...+101+4101 (100 times)
100 x 101
So, S= > 5050 , i.e., the sum = 5050.
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64 MATHEMATICS

We will now use the same technique to find the sum of the first n terms of an AP :

a,a+d,at+2d,...

The ath term of this APis a + (n - 1) d. Let S denote the sum of the first terms
of the AP. We have

S=at+(atd)+(at+2d)+...+[a+(m---1)d] (1)
Rewriting the terms in reverse order, we have

S=[a+(n--1)d]+flat+(n---2)d]+...+(a+d)+a (2)
On adding (1) and (2), term-wise. we get °

[2a+ (vn --Dd)+[2a+ (--)d)+...4+ [24+ (n -- 1d) +[2at+ @ -- 1d]
- ee _
28 = n times .
o, 2S=n[2a+(n--1)d] (Since, there are m terms) . _
Or, S= 5 [2a + (m-- 1) d] . . . -
So, the sum of the first 1 terms of an AP is given by
n
S= 3 [2a + (2 --1) d]
. . 7 . n :
We can also write this *) S= 2 [a+ a+(n-1)d]
n
1.e., S= 3 (a+a,) (3)
Now, if there are only n terms in an AP, then a, = 1, the last term.
From (3), we see that ~
. n
7 S=- (+l) (4)
. . 2

This form of the result is useful when the first and the last terms of an AP are
given and the common difference is not given.

Now we return to the question that was posed to us in the beginning. The amount
of money (in Rs) in the money box of Shakila's daughter on 1st, 2nd, 3rd, 4th birthday,
..., were 100, 150, 200, 250, . . ., respectively.

This is an AP. We have to find the total money collected on her 21st birthday, i.e.,
the sum of the first 21 terms of this AP.

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ARITHMETIC PROGRESSIONS 65
Here, a= 100, d=50 and n =21. Using the formula :

n

S= gat @-D d],

21 21
we have S= 3 [2100+ 21-1) x50) = 5 [200 + 1000]

21

= 5 x 1200 = 12600

So, the amount of money collected on her 21st birthday is = 12600.

Hasn't the use of the formula made it much easier to solve the problem?

We also use Sin place of S to denote the sum of first n terms of the AP. We
write S,, to denote the sum of the first 20 terms of an AP. The formula for the sum of
the first n terms involves four quantities S, a, d and n. If we know any three of them,
we can find the fourth. - .

Remark : The nth term of an AP is the difference of the sum to first n terms and the
sum to first (n - 1) terms of it,ie.,a,=S,-S, oj."

Let us consider some examples. ~ .

Example 11 : Find the sum of the first 22 terms of the AP: 8, 3, -2,...
Solution : Here, a= 8, d=3-8=-5, n= 22.
f ;
We know that )
MN n .
S = at (nda
22
Therefore, S=> [16 + 21(-5)] = 11116 - 105) = 11(-89) = - 979
So, the sum of the first 22 terms of the AP is - 979.
Example 12 : If the sum of the first 14 terms of an AP is 1050 and its first term is 10,
find the 20th term.
Solution : Here, S,, = 1050, n= 14, a=10.
n
As S.= 724+ (n-Dd],
14
so, 1050= 7 120+ 134] = 140+ 91d
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66 MATHEMATICS
Le., 910= 91d
Or, d= 10
Therefore, Ay) = 10 + (20-1) x 10 = 200, i.e. 20th term is 200.
Example 13 : How many terms of the AP: 24, 21, 18, .. . must be taken so that their
sum is 78?
Solution : Here, a = 24, d= 21 --24=-3, S = 78. We need to find n.
n
We know that S= i244 (n--Nd] :
n n ;
So, 78= 5[48+(2-D-3)] = 5[51-37] |
or 3n2 -51n + 156 = 0 . : .
or nw -17In+52= 0 . -
or (n--4)(n - 13) = 0 - _
or n=4 or 13. . / :
Both values of n are admissible. So, the number of terms is either 4 or 13.
Remarks:

1. In this case, the sum of the first 4 terms = the sum of the first 13 terms = 78.

2. Two answers are possible because the sum of the terms from 5th to 13th will be
zero. This is because a is positive and d is negative, so that some terms will be
positive and some others negative, and will cancel out each other.

Example 14 : Find the sum of :
(i) the first 1000 positive integers (ii) the first n positive integers
Solution : 4
(i) LetS=1+2+3+...+ 1000
. n
Using the formula S_ = 54 +1) for the sum of the first n terms of an AP, we
have
° 1000
: Si000 = > (1+ 1000) = 500 x 1001 = 500500
So, the sum of the first 1000 positive integers is 500500.
(i) LetS =1+2+4+3+...4+n
Here a = | and the last term / is n.
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ARITHMETIC PROGRESSIONS 67

n(l+n) n(iat+l

Therefore, S.= > or S = a

So, the sum of first n positive integers is given by

n(n +1)
S= 32

Example 15 : Find the sum of first 24 terms of the list of numbers whose nth term is

given by .
a,=3+2n

Solution : ; .

As a,=342n,

So, a,=34+2=5 , So
a=342x2=57
a,=3+2x3=9 cs _

List of numbers becomes 5, 7,9, 11,... .

Here, 7-5=9-7=11-9=2 and s0 on.

So, it forms an AP with common difference d = 2.

To find S,,, we have ne 24, a=5, d=2..

SS .

Therefore, S,,= = [2x5 +(24-1)x2] = 12[10+ 46] = 672

So, sum of first 24 terms of the list of numbers is 672.

Example 16 : A manufacturer of TV sets produced 600 sets in the third year and 700

sets in the seventh year. Assuming that the production increases uniformly by a fixed

number every year, find :
(i) the production in the Ist year (ii) the production in the 10th year

(iii) the total production in first 7 years

Solution : (i) Since the production increases uniformly by a fixed number every year,

the number of TV sets manufactured in Ist, 2nd, 3rd, ..., years will form an AP.

Let us denote the number of TV sets manufactured in the nth year by a...

Then, a,= 600 and a, = 700

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