Shubharambha OPT Mathematics 10 Final for CTP 2077

or, taN+– 3d = 4
2d 9

or, 36 + 27d = 176 – 8d

or, 35d = 140

∴ d = 4

Again tn = a + (n – 1)d
or, 44 = 4 + (n + 2 – 1) 4

or, 40 = (n + 1)4

∴ n = 9

Hence, no of A.M. (n) = 9



Alternative Method:

(n 3rd mean = 4
– 1)th mean 9

or, a +a(n+ 3d d = 4
– 1) 9

or, 4 4 + 3d = 4
+ (n – 1)d 9

or, 36 + 27 d = 16 + 4 nd – 4d
or, 20 = 4nd – 31d

or, 20 =d
4n – 31

Again, tN = a + (N – 1)d

or, 44 = 4 + (n + 2 – 1) × 20
4n – 31

or, 40 = (n + 1) 20
4n – 31

or, 2 = n + 1
4n – 31

or, 8n – 62 = n + 1
or, 7n = 63
∴ n = 9
Hence, no. of A.M. (n) = 9.

Infinity Optional Mathematics Book - 10 51

Exercise 1.7

Section 'A'

1. Find the arithmetic mean between the following two numbers:

(a) 4 and 16 (b) – 35 and 55

(d) (3x+2y) and (5x-4y) (e) (a+b)2and (–a+b)2

2. From the following given terms of A.P. , find the indicated terms:
(a) 1st terms = 7, 3rd term =13, 2nd term = ?
(b) 5th term = 32, 7th term = 48, 6th term = ?
(c) 9th term = –81, 11th term= –50, 10th term =?
(d) 65th term= 251, 67th term = 321, 66th term =?
3. Find the value of x and y in the following A.P.:

(a) 4, x, 10 (b) 15, y, 29 (c) 5, x, 11, y

(d) –5, y, 5,x (e) x, 30, 3x (f) 7x, x2, – 6

(g) 5x, 3, x2 (h) y, 275, 1y (i) 4x-1, 2x-5, 6x+1

Section 'B'

4. (a) Insert 3 arithmetic mean between 3 and 19.

(b) Insert 4 arithmetic mean between 7 and -13.

(c) Insert 5 arithmetic mean between 1 and 133.
3

5. (a) Find the value of a, b, and c when 2, a, b, c, 14 are in A.P.

(b) Find the values of p, q, r and s when –7, p, q, r, s, 18 are in A.P.

(c) Find the values of x, y and z when –5, x, y, z, 15 are in A.P.

(d) Find the values of x and y when 7, x, y, 9 are in A.P.

6. (a) There are 5 arithmetic means between x and y. If the first mean and 4th
mean are 4 and 10 respectively, then find the values of x and y.

(b) There are 6 arithmetic means between a and b. If the 2nd mean and 4th
mean are 9 and 15 respectively then find the values of a and b.

52 Infinity Optional Mathematics Book - 10

(c) There are 6 arithmetic means between p and q. If the second mean and
last mean are 8 and 28 respectively, find the values of p and q.

7. (a) There are 5 arithmetic means between 5 and b. If the third mean is 20,
then find the value of b and other means.

(b) There are 6 A.M's between 10 and k. If the last mean is –10 , then find
the value of k and the other means.

8. (a) Some arithmetic means are inserted between 10 and –14. If the 3rd mean
is –2, find the number of arithmetic means.

(b) The 4th mean between 5 and 35 is 25. Find the number of arithmetic
mens.

9. (a) There are n arithmetic means between 3 and 90. The ratio of 10th mean
to the 5th mean is 11:6. Find the value of n.

(b) There are n arithmetic means between 14 and –22. The ratio of 2nd mean
to the second last mean is –3:7. Find the value of n.

(c) There are y arithmetic means between 5 and 50. The ratio of 1st mean to
the (y-3)th mean is 1:3, find the value y.

(d) There are n arithmetic means between 100 and 2. The ratio of (n –2)th
mean to the (n – 6)th mean is 1:2, find the value of n.

Infinity Optional Mathematics Book - 10 53

Sum of the Arithmetic Series

Sum of first n terms of an A.S.

Let us consider the numbers in A.P. be

1, 3, 5, 7, 9, 11, 13, 15, 17, 19

Here, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100

Hence, the result obtained by adding the terms of an A.P. is known as sum of A.P.
In A.P, sum of n terms is denoted by Sn.
S10 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
or, S10 = 19 + 17 + 15 + 13 + 11 + 9 + 7 + 5 + 3 + 1
or, 2S10 = 10 × 20

or, S10 = 10 × 20
2

∴ S10 = 100
Similarly, when a, d, n and l are the first term, common difference, number of terms
and last term of an A.P. respectively, then,

a, a + d, a + 2d, ...., l – 2d, l – d, l are in A.P.

If Sn is the sum of the first n terms of A.P.
Then, Sn = a + (a + d) + (a + 2d) + … (l – 2d) + (l – d) + l … (i)
Again, writing the terms in reverse order,

Sn = l + (l – d) + (l – 2d) + … + (a + 2d) + (a + d) + a … (ii)
Adding (i) and (ii), we get

2Sn = (a + l) + (a + l) + (a + l) + … (a + l) + (a + l) + (l + a)
Since, there are altogether n terms.

So, 2Sn = n(a + l)

∴ Sn = n (a + l)
2

Again, l = tn = a + (n – 1)d

So, Sn = n [(a + a + (n – 1)d]
2

∴ Sn = n [2a + (n – 1)d]
2

Hence, the sum of first n terms when first term (a), last term (l) and number of

54 Infinity Optional Mathematics Book - 10

terms (n) is given by

Sn = n (a + l)
2

When first term (a), number of term (n) and common difference (d) are given, then
the sum of first n terms of an AP is given by:

Sn = n [2a + (n – 1)d]
2


Note : We can suppose the three numbers in A.P. as a – d, a and a + d. Where a–
d, a and a + d are the first term, second term and third term respectively
and d is the common difference.

WORKED OUT EXAMPLES

1. Find the sum of the following series:

(a) 4 + 7 + 10 + … to 20 terms.

(b) 2 – 9 – 20 – … –130
Solution:

(a) 4 + 7 + 10 + … to 20 terms

Here, first term (a) = 4

Common difference (d) = t2 – t1 = 7 – 4 = 3
No. of terms (n) = 20

Now, sum of first n terms of an A.P. is

Sn = n [2a + (n – 1)d]
2

= 220[2 × 4 + (20 – 1) 3]

= 10 [8 + 57]

= 650

(b) 2 – 9 – 20 – … –130

Here, First term (a) = 2

Common difference (d) = t2 – t1 = – 9 – 2 = – 11

Last term (tn) = l = – 130

Now, tn = a + (n – 1) d

or, – 130 = 2 + (n – 1) (– 11)

Infinity Optional Mathematics Book - 10 55

or, –132 + 1 = n
–11

∴ n = 13

Now, sum of the series (Sn ) = n ( a + l)]
2

= 13 (2 – 130)
2

= 13 × (– 128)
2

= – 13 × 64

15 = – 832

Σ2. Find the value of: (n + 2)
Solution:
15 n=1

Σ Here, (n + 2) = (1 + 2) + (2 + 2) + (3 +2)+ …+ (15 + 2)
= 3 + 4 + 5 + …+ 17
n=1


Here, first term (a) = 3

common difference (d) = t2 – t1 = 4 – 3 = 1
No. of terms (n) = 15

last term (l) = 17

Now, sum (sn) = n (a + l)
2

= 15 (3 + 17)
2

= 15 × 20
2

= 150

3. How many terms of the series 24 + 20 + 16 + … must be taken so that

the sum may be 72? Explain the double answer.

Solution: Here, the given series is

24 + 20 + 16 + …

first term (a) = 24

common difference (d) = t2 – t1 = 20 – 24 = – 4

sum of the series (sn) = 72
Now, sum (n2S n )[2=×n224[2+a + (n – 1) d]
or, 72 =
(n – 1) (–4)]

56 Infinity Optional Mathematics Book - 10

or, 144 = n (48 – 4n + 4]

or, 144 = n(48 – 4n + 4)

or, 144 = 52n – 4n2
or, 4n2 – 52n + 144 = 0
or, n2 – 13n + 36 = 0

or, n2 – 9n – 4n + 36 = 0

or, n (n – 9) – 4 (n – 9) = 0

or, (n – 9) (n – 4) = 0

Either, n – 9 = 0 or, n – 4 = 0

∴ n = 9 ∴ n = 4

∴ The number of terms are 9 or 4

Here, the sum of the series having 4 terms is 24 + 20 + 16 + 12 = 72

The sum of the series having 9 terms is

24 + 20 + 16 + 12 + 8 + 4 + 0 – 4 – 8 = 72

Here, sum of the last five terms is zero.

∴ Sum of first 4 terms = sum of first 9 terms.

4. The sum of the first 4 terms of an A.P. is 26 and the sum of the first 8
terms is 100, find the sum of the first 20 terms.

Solution:

Here, sum of first 4 terms (Sn) = 26

or, n2 [2a + (n – 1) d] = 26

or, 4 [2a + (4 – 1)d] = 26
2

or, 2(2a + 3d) = 26

or, 2a + 3d = 13

or, a= 13 – 3d …(i)
2

Sum of the first 8 terms (S8) = 100

or, 82 [2a + (8 – 1)d] = 100

or, 4 (2a + 7d) = 100

or, 2a + 7d = 25

or, a= 25 – 7d …(ii)
2 Infinity Optional Mathematics Book - 10 57

From equation (i) and (ii), we get

13 – 3d = 25 – 7d
2 2
or, 13 – 3d = 25 – 7d

or, 4d = 12

∴ d=3

Putting the value of d in equation (ii), we get

a= 13 – 3d = 13 – 3 × 3 = 4
2 2 2
∴ a=2

Now, sum of first 20 terms

S20 = 20 [2a + (20 – 1)d]
2

= 10 (2 × 2 + 19 × 3]

= 10 (4 + 57)

= 610

5. Find the sum of the numbers from 101 to 256 which are divisible by 7.

Solution:
Here, the numbers between 101 to 256 which are divisible by 7 form an AP.

i.e. 105 + 112 + 119 + …+ 252 are in A.P.

Here, first term (a) = 105 Note:

common difference (d) = 7 7)101 (14 7) 256 (14

last term (tn) = (l) = 252 –7 –21
Now, tn = a + (n – 1) d
or, 252 = 105 + (n – 1) 7 31 46

or, 1477 = n – 1 –28 –42

3 4

1st term =101+(7–3) last term = 256–4

= 105 = 252

∴ n = 22

Now, sum of the series (Sn) = n (a + l)
2

S22 = 22 (105 + 252) = 11 × 357 = 3927
2
6. Find the three numbers in AP such that their sum is 18 and the sum

of their product in pairs is 92.

Solution:

Let, the three numbers in AP be

58 Infinity Optional Mathematics Book - 10

a – d, a and a + d.
By first case,
a – d + a + a + d = 18
∴ a=6
By second case,
(a – d ) a + a (a + d) + (a – d) (a + d) = 92

or, a2 – ad + a2 + ad + a2 – d2 = 92

or, 3a2 – d2 = 92

or, 3 × 36 – d2 = 92 [ a = 6]

or, d2 = 16

∴ d = ± 4

When a = 6 and d = 4, the three number are

a – d = 6 – 4 = 2

∴ a = 6

Again, a + d = 6 + 4 = 10

When a = 6 and d = – 4, the three numbers are

a – d = 6 –(– 4) = 10

∴ a = 6

Again, a + d = 6 – 4 = 2

Hence, the three numbers in AP are 2, 6, 10, or 10, 6, 2.

Exercise 1.8

Section 'A'

1. Find the sum of the following series:

(a) 3+7+11+ … 15 terms (b) 50+45+40+ … 10 terms.

(c) –15 –12 –9 – … 20 terms. (d) 3 + 8 + 13 + ...... + 58

(e) 13 + 2 – 9 – … – 144

2. Find the value of : 25 20

15 (b) Σ (n – 4) (c) Σ (n + 1)

(a) Σ (3n + 1) n=1 n=2

n=1

3. (a) The first term and the common difference of an arithmetic series are 20
and 5 respectively. Find the sum of its first 10 terms.

(b) Find the sum of first 20 terms of an arithmetic series whose first term is
13 and common difference is –4.

(c) The last term of an arithmetic series of 20 terms is 195 and the common
difference is 5. Calculate the sum of the series.

4. (a) Find the first term of arithmetic progression whose common difference is
4 and sum of its first 20 terms is 820.

Infinity Optional Mathematics Book - 10 59

(b) The sum of first 30 terms of an arithmetic progression is –2310 and the
common difference is –6 . Find the first term.

5. (a) The sum of first 10 terms of the series is 110. If the 10th term is 20, then
find the first term.

(b) An arithmetic series with 30 terms has the sum 2235. If the first term is
2, then find its last term.

6. (a) The sum of first 20 terms of an A.P. is 820. If the first term is 3, then find
the common difference.

(b) The sum of first 50 terms of an A.P. is –2400. If the first term is 50, then
find the common difference.

7. (a) Find the number of terms in arithmetic series whose first term is 4,
common difference is 3 and sum of the series is 246.

(b) Find the number of terms is an arithmetic progression which has its first
term 16, common difference 4 and the sum 120.

(c) The first term and last term of an A.P. are 3 and 90 respectively. If the
sum of the progression is 1395, then find the number of terms and the
common difference.

(d) The first term, last term and the sum of the arithmetic series are 2,
29 and 155 respectively. Find the number of terms and the common
difference.

8. (a) How many terms of the series 30+27+24+ …must be taken so that the
sum of the series may be 162 ? Explain the double answer.

(b) How many terms of the series 20+18+16+ …must be taken so that the
sum of the series may be 110 ?Explain the double answer.

Section 'B'
9. (a) If the 5th term and 9th term of an arithmetic series are 10 and 18

respectively, find the sum of first 20 terms.
(b) If the 3rd term and the 6th term of an arithmetic series are 34 and 19

respectively , find the sum of first 10 terms.
10. (a) The 5th term of an A.P. is 15 and the sum of the first 10 terms is 165.

Find the sum of the first 20 terms.
(b) The sum of the first 12 terms of an arithmetic series is 168 and the 9th

term is 19. Find the sum of the first 25 terms.
11. (a) The sum of the first two terms and the sum of the first five terms of an

arithmetic series are 6 and 30 respectively. Find the sum of the first 10
terms.
(b) The sum of the first 3 terms and the sum of first 8 terms of an arithmetic
series are 21 and 136 respectively. Find the sum of first 15 terms.

60 Infinity Optional Mathematics Book - 10

(c) The sum of first 15 terms is 45 and that of 19 terms is 95. Find the
number of terms of the series from the first having the sum 290.

12. (a) The sum of the three consecutive terms of an A.P. is 54 and the product
of two extremes is 275. Find the three terms.

(b) The sum of three terms of an A.P. is 21 and their product is 280. Find the
three terms.

(c) The sum of three terms of an AP is 24 and the third term is seven times
the first term. Find the three terms.

(d) The sum of three terms of an arithmetic series is 12 and the sum of their
squares is 50 Find the three terms.

(e) The sum of three terms of an arithmetic series is15 and the sum of the
squares of its first and the third is 58. Find the three terms.

(f) There are three numbers in A.P. Difference between the first and the
third is 8 and the product of these two is 33. Find the three numbers.

13. (a) Find the sum of all the numbers from 50 to 195 which are exactly divisible by 4.

(b) Find the sum of all the numbers from 250 to 780 which are exactly
divisible by 11.

(c) Find the sum of all two digit numbers in A.P. which are the multiple of 7.

(d) Find the sum of all the three digits numbers in A.P. which are the
multiple of 15.

(e) Find the sum of the first 50 natural numbers.

(f) Find the sum of even integers between 101 and 503.

(g) Find the sum of odd integers between 30 to 100.
14. (a) Find the sum of first 15 terms of an A.P. whose 8th term is 45.
(b) If the 10th term of an A.P. is 200, then find the sum of first 19 terms.

(c) The sum of first three terms of an arithmetic progression is 21. If the
sum of first two terms is subtracted from the third term, them it would
be 9. Find the three terms of the series.

(d) In an A.P., the 6th terms is equal to 3 times the 4th term and the sum of
the first three terms is –12. Find the A.P.

(e) The sum of the 4th and 8th terms of an A.P. is 24 and the sum of the 6th
and 10th term is 34 of the progression. Find the sum of the 1st 10 terms.

15. (a) A man repays a loan of Rs. 24750 by paying Rs. 100 in the first month
and then increases the payment by Rs. 50 every month. How long will it
take him to clear the loan ?

(b) A man saved Rs. 70200 in 3 years. In each month after the first, he saved
Rs. 100 more than he did in the preceding month. How much did he save
in the first month?

Infinity Optional Mathematics Book - 10 61

Geometric Progression (G.P.)

Let us consider the two examples.
(a) 2 + 4 + 8 + 16 + 32 + …
Here the numbers are increased in the multiple of 2.

(b) 8 –4 + 2 – 1 + 21 – 1 …
4

Here the numbers are decreased by the multiple of –21.

In the above two examples, the numbers are increased or decreased by a constant
number called common ratio. They are called geometric progression. A geometric
progression (G.P) is also known as geometric series or sequence (G.S.)
Hence a sequence or series of numbers that increased or decreased with a constant
ratio is called a geometric progression or geometric series (or sequence).

The constant ratio is called common ratio and is denoted by r.

common ratio(r) = succeeding term
preceeding term

i.e. r= t2 = t3 = t4 …
t1 t2 t3

General term or nth term of a G.P. or (G.S.)
In a G.P, the following notations are used

First term = t1 = a
common ratio = r

last term or nth term = tn = l
sum of first n terms = Sn
Let us consider a G.P.

3 + 9 + 27 + 81 + 243 +…

Here, common ratio = t2 = 93 = 3.
t1

first term (t1) = 3 = 3.3º = a.rº = ar1 – 1
Second term (t2) = 9 = 3 × 3 = 3 × 31 = a.r1 = ar2 – 1
third term (t3) = 27 = 3 × 9 = 3 × 32 = a.r 3 – 1
Fourth term (t4) = 81 = 3 × 27 = 3 × 33 = a × r 4 – 1

………………………………………………

nth term (tn) = l = a × rn – 1 = arn – 1
Hence, the nth term or general term of an GP is (tn = arn – 1)

62 Infinity Optional Mathematics Book - 10

Note: For the first term, n = 1

∴ t1 = ar1 – 1 = arº = a
For second term, n = 2

∴ t2 = ar2 – 1 = ar
For third term, n = 3

∴ t3 = ar3 – 1 = ar2
For fourth term, n = 4

t4 = ar 4 – 1 = ar3.
Similarly: 6th term, t6 = ar5
10th term, t10 = ar9 and so on

WORKED OUT EXAMPLES

1. In the given geometric sequence: 3, 6, 12, 24,…find

(i) common ratio (r)

(ii) the general term (tn) iii) next three terms.
Solution:
Here, the given geometric sequence is

3, 6, 12, 24,…

(i) common ratio (r) = t2 = 6 =2
t1 3

(ii) the general term (tn) = ar n – 1 = 3.2n – 1
(iii) 5th term (t5) = 3.25 – 1 = 3.24 = 3 × 16 = 48
6th term (t6) = 3.26 – 1 = 3.25 = 3 × 32 = 96
7th term (t7) = 3.2 7 – 1 = 3.26 = 3× 64 = 192
2. Find the 10th term of the GP: 2 – 2 2 + 4 – 4 4 + …

Solution:

Here,the given G.P. is 2 – 2 2 + 4 – 4 2 + …

Here, first term (a) = 2

common ratio (r) = t2 = –2 2 =– 2
t1 2

Now, 10th term (t10) = arn – 1

= 2 (– 2)10 – 1

= 2(– 2)9

Infinity Optional Mathematics Book - 10 63

=–2×2×2×2×2 2
= – 32 2

3. Find the number of terms of the series: 116 – 81 + 14 – 12 …+ 16
Solution:

Here, first term (a) = 116 –1

common ratio (r) = t2 = 8 = –16 = – 2
t1 1 8
last term (tn) = 16 16

Now, tn = arn – 1

or, 16 = 1 × (– 2)n – 1
16

or, 256 = (– 2)n – 1
or, 28 = ( –2)n – 1
or, (– 2)8 = (– 2)n – 1
or, n – 1 = 8
∴ n=9

4. Find the common ratio of a G.P. whose first term is 116 and 11th term
is 2.

Solution: Here, first term (a) = 116

11th term (t11) = 2
Now, t11 = ar10

or, 2= 1 r10
16

or, 32 = r10
or, 25 = r10
or, ( 2)10 = r10
∴ r = 2

64 Infinity Optional Mathematics Book - 10

5 . The 2nd term and 5th term of a GP are 4 and 256 respectively. Find the
10th term.

Solution:

Here, 2nd term (t2) = 4
or, ar = 4 …(i)

5th term (t5) = 256
or, ar4 = 256 …(ii)

Now, dividing equation (ii) by equation (i), we get

aarr4 = 256
4
or, r3 = 64

or, r3 = 43

∴ r = 4

Putting the value of r in equation (i), we get

4a = 4

∴ a=1
Now, 10th term (t10) = ar9 = 1 × 49 = 262144.

Exercise 1.9

Section 'A'

1. Determine with reason, which of the following are in geometric
progression:

(a) 3, 6, 12, 24, … (b) –100 + 50 –1 + …

(c) 2 + 2 + 2 2 + … (d) 116 + 1 + 1 + …
8 4
1
(e) –9 +3 –1 + 3 –…

2. From the following geometric progressions given below, find: (i) the

common ratio, (ii) the general terms (tn) (iii) the next three terms.

(a) 2, 6, 18, … (b) –1 – 1 – 13, …
(c) –1 + 2 –2 + … 27 9
(d) 80, 20, 5, …

(e) –23 + 3 – 6 + …
3. Find the first 3 terms of the following geometric progressions:

(a) First term (a) =2, common ration (r) = 3.

Infinity Optional Mathematics Book - 10 65

(b) First term (a) = 2, common ratio (r) = 2
(c) First term (a) = –91,
(d) First term (a) = 16, common ratio (r) = – 3.

common ratio (r) = 1
2

(e) First term (a) = 8 , common ratio (r) = 32.
27

4. Find the 4th term and 7th term of the following geometric sequences:

(a) –1 – 1 – 1 …
16 8 4

(b ) First term = 3 and common ratio = 2
2 3

(c) First term = 5 and common ratio = 3

5. Find the first term from each of the following geometric progressions:

(a) Common ratio = 2 and 6th term = 64

(b) Common ratio = – 2 and 11th term = 32

(c) Common ratio = 1 and 7th term = 1
3 27

6. Find the common ratio in each of the following sequences:

(a) First term = 2 and 10th term = 1024.

(b) First term = 3 and 7th term = 32
2 243

(c) First term =1 and 12th term = – 32 2.

(d) First term = 64 and 7th term =1
729

7. Find the number of terms in each of the following sequences:

(a) First term = 2, common ratio = 2 and last term = 512

(b) First term = – 3 , common ratio = – 3 and last term = 81 3

(c) First term = – 1 , common ratio = – 3 and last term = 729
9
243 81 27 32
(d) 32 + 16 + 8 + … + 243

(e) 1–215 + 1 – 1 + … + 625
125 5

8. Which term of the series:

(a) 3 2 + 6 +6 2 + … is 96 2 ?

(b) 2 – 4 + 16 … is 128 ?
3 9 27 2187

(c) 1 + 1 + 1 … is 81 ?
81 27 9

66 Infinity Optional Mathematics Book - 10

9. (a) Is 64 a term of the series 1 + 1 + 1 … ?
64 32 16

(b) Is – 16 2 a term of the series – 1 – 2 – 2 …?

(c) Is 64 a term of the G.P. whose first term is 27 and common ratio is 23?
27
1 1 1
(d) Is 512 a term of the series 2 + 4 + 8 …?

10. (a) Find the 8th term of the geometric progression whose first term is 1 and
2

3rd term is 2.

(b) Find the 5th term of the geometric series whose common ratio is 2 and

8th term is 2 2.

11. Find the terms indicated as follows.

(a) 3rd term = 8 7th term = 128, 10th term = ?
2nd term = 91, 8th term = ?
(b) 5th term = 3 6th term = 2, 20th term = ?

(c) 4th term = 1 ,
2

Section 'B'

12. (a) If 3rd term and 6th term of a geometric series are 1 and 16 respectively,
4

then which term of the series is 256?

(b) Which term of the geometric series is 2 2 , whose 3rd term is 1 and 7th
22
term is 2 ?

(c) Is 64 a term of the geometric series whose 4th term is 1 and 6th term is 1
64

?

13. (a) The fifth term of a geometric progression is 8 times the second term and

the 7th term is 128. Find the series.

(b) The 8th term of a geometric series is 1 times 5th term and 2nd term is 9.
27
Find the 10th term.

(c) The 4th term of a geometric series is square of its second term and the
first term is – 3. Find its 10th term.

Infinity Optional Mathematics Book - 10 67

Geometric Mean (G.M.)

Consider the following geometric sequences:
(i) 2G, 6.M, 1.8 (ii) 91, 31G, .1M, 3s' (iii) 2, 4, 1G6,.M32,s6' 4, 128

In the above G.P., the terms except the first term and last term are geometric means.

Thus the terms to be placed between first term and last term of a geometric
progression are called geometric means.

Let us consider the three term a, m and b of a G.P. then, m is the G.M of a and b.

When a, m, b are in G.P. then their common ratio are equal.

i.e. m = b
a m
or, m2 = ab

∴ m = ab

Hence, G.M. between a and b = ab

The geometric means between first term and last term are denoted by m1, m2, m3,
m4…..where m1 = t2, m2 = t3 , m3 = t4 and so on.

WORKED OUT EXAMPLES

1. Find the geometric mean between 9 and 25.

Solution:

Here, the two given numbers are 9 and 25.

Now, G.M. between 9 and 25 = 9 × 25

= 3 × 5

= 15

2. Find the 10th term of a G.P. whose 9th term is 1 and 11th term is 9.
9
Solution:

Here, 9th term (t9) = 1
9
11th term (t11) = 9

Here, t10 is G.M between t9 and t11. So,

t10 = t9 × t11 = 1 × 9 =1
9

68 Infinity Optional Mathematics Book - 10

3. Find the value of x and y when 1 , x, 4, y are in G.P.
Solution 4

Here, the gthiveeGn .GM.Pb.e41tw, exe, n4,41y and 4
Here, x is

So, x = 1 × 4 = 1
4

Again, 4 is the G.M between x and y.

So, 4 = xy

or, 4 = 1 × y

Now, squaring on both sides

y = 16

Hence, x = 1 and y = 16

4. Insert 4 geometric means between 4 and 128.
Solution: Here, first term (a) = 4

last term (tn) = 128
No. of terms (n) = 4 + 2 = 6

Now, tn = arn – 1
or, 128 = 4 × r 6 – 1

or, 32 = r5

or, 25 = r5

∴ r = 2

Now, 1st mean (m1) = t2 = ar = 4 × 2 = 8
2nd mean (m2) = t3 = ar2 = 4 × 22 = 4 × 4 = 16
3rd mean (m3) = t4 = ar3 = 4 × 23 = 4 × 8 = 32
4th mean (m4) = t5 = ar4 = 4 × 24 = 4 × 16 = 64

Hence, the four G.MS' are 8, 16, 32, 64.

Note: In a G.P.

Terms t1 t2 t3 t4 … tN – 3 tN – 2 tN – 1 tN
Means a
m1 m2 m3 … mn– 2 mn – 1 mn
ar ar2 ar3 … tN
r3 tN tN
r2 r

Infinity Optional Mathematics Book - 10 69

5. There are n G.MS' between 16 and 21463. The ratio of 4th mean to the
(n – 1)th mean is 4:9. Find n. 27

Solution: 16 Alternative Method:
27
Here, First term (a) = 4th mean 4
– 1)th mean 9
243 Here, (n =
16
Last term (tN) =

No. of term (N) = n + 2 or, ar4 1 = 4
4th mean : (n – 1)th mean = 4 : 9 arn – 9

or, (n 4th mean = 4 or, 49r5 = rn ................. (i)
– 1)th mean 9

or, 3rd5lthasttertmerm = 4
9
ar4 Again,
tN 4
or, = 9 tN = arN – 1

r2 r2
ar4 ×
or, tN = 4 or, 243 = 16 rn + 2 – 1
9 16 27

16 × r6 4 or, 243 × 27 = rn + 1
27 9 16 × 16
or, 243 =

r6 1=694 243 × 27 or, 3 8 rn . r or, 3 8 94r5 × r
16 × 16 2 2
or, × = =

or, r6 = 36 or, 3 6 ∴ r = 3
26 2 2
= r6

or, r6 = 36 Putting the value of r in eqn (i), we get
2

∴ r= 3 9 3 5 3n
2 4 2 2
Again, tN = arN – 1 × =

243 16 3 n+2–1 3 7 3 n
16 27 2 2 2
or, = × or, = ∴n=7

or, 243 = 16 × 3 n+1 Hence, no. of G.M. = 7.
16 27 2

or, 243 × 27 = 3 n+1
16 × 16 2

or, 38 = 3 n+1
28 2

or, 3 8 3 n+1
2 2
=

or, 8 = n + 1

∴ n=7

Hence, no. of G.M. = 7

70 Infinity Optional Mathematics Book - 10

Exercise 1.10

Section 'A'

1. Find the geometric mean between the following two numbers:

(a) 4 and 16 (b) – 25 and – 36 (c) 1 and 81
(d) 25 and 16 (e) 27 and 3 9

4 625 82

2. From the following given terms of a G.P, find the indicated terms:

(a) 1st term = 4 and third term = 16, 2nd term = ?

(b) 5th term = 1 and 7th term = 9, 6th term= ?
9

(c) 6th term = 1 and 10th term = 1, 8th term =?
64 4

(d) 2nd term = 25 and 8th term = 25, 5th term=?
6 16

3. Find the values of x and y in each of the following G.P.:

(a) 2, x, 8 (b) 1 , y, 49 (c) 3, x, 12, y
94

(d) x–5, x + 1, x + 5 (e) 3y, y + 6, 3y + 8

Section 'B'

4. (a) Insert 3 geometric means between 2 and 32.

(b) Insert 4 geometric means between 2 and – 8.

(c) Insert 5 geometric means between 4 and 81 .
9 16

(d) Insert 3 geometric means between 1 and 16.
16

5. (a) Find the values of x,y and z when 1, x, y, z, 9 are in G.P.
9

(b) Find the value of p,q,r and s when – 2, p,q,r,s – 8 are in G.P.

(c) Find the values of a, b, c, d and e when 243, a, b, c, d, e, 1 are in G.P.
3

Infinity Optional Mathematics Book - 10 71

Section 'C'

6. (a) There are 5 geometric means between x and y. If the 2nd mean and 5th
mean are 27 and 1 respectively, then find the values of x and y.

(b) There are 8 geometric means between p and q . If the 1st mean and 7th
mean are 1 and 2 respectively, then find the values of p and q.
32

7. (a) There are 4 geometric means between 4 and k. If the 2nd mean is 36, then

find the value of k and the other means.

(b) There are 6 geometric means between – 2 and x. If the 4th mean is –4 2
, then find the value of x and the other means.

8. (a) Find the number of geometric means between 1 and 64, where the
64

common ratio is 4.

(b) Some geometric means are inserted between 2 and 16. If the third mean
is – 4 2, find the number of G.M.

(c) The 5th mean between 1 and 16 is 1 . Find the number of geometric
32

means.

9. (a) There are n geometric means between 2 and 64. If 1st mean: 4th mean =
1:8, find n.

(b) There are n geometric means between 1 and 81. If the ratio of 3rd mean
81

to the last mean is 1:81, then find the value of n.

(c) There are n geometric means between 1 and 4096. If the ratio of (n – 2)th
mean to the (n – 5)th mean is 8:1, then find the value of n.

10. If ax = by = cz and a, b, c are in G.P. then show that 1 + 1 = 2
x z y

72 Infinity Optional Mathematics Book - 10

Sum of the Geometric series

Sum of first n terms of a G.S.

Let us consider the numbers in G.P.

2, 4, 8, 16, 32, 64, 128.

then 2 + 4 + 8 + 16 + 32 + 64 + 128 = 254

Hence, the result obtained by adding the terms of a G.P. is known as sum of G.P. In
GP, sum of n terms is denoted by Sn.

When a, r, n and l are first term, common ratio, number of terms and last term of a
G.P. respectively then the G.P. is

a, ar, ar2, ar3 … l , l , l , l
r3 r2 r

The sum of first n terms

Sn = a + ar + ar2 + ar3 + … l + l + l + l…(i)
r3 r2 r

Multiplying both sides of equation (i) by r, we get

rSn = ar, ar2, ar3 .... + … l + l + l + l + lr, …(ii)
r3 r2 r

Now, subtracting equation (i) from equation (ii), we get

rSn = ar + ar2 + ar3 + … l + l + l + lr
r2 r

Sn = a + ar + ar2 + .................. + l + l + l +l
r3 r2 r
(–) (–) (–) (–) (–) (–) (–) (–)

∴ Sn = lr – a Sn (r – 1) = – a + lr
r–1

where, l = arn – 1

Sn = arn – 1 × r – a
r–1

= arn – 1 + 1 – a = arn – a
r–1 r–1

∴ Sn = a(rn – 1)
r–1

Note:arare,taheafnirdsta,rsceaconnbde asnudp p tohsierdd as three numbers in G.P. wthheecroemarm, aonanradtiaor.
term respectively and r is

Infinity Optional Mathematics Book - 10 73

WORKED OUT EXAMPLES

1. Find the sum of the following series

(a) 2 – 4 + 8 – 16 to 10 terms.
S olut(ibo)n : 19 + 1
3 + 1 + …+ 9

(a) Here, the given series is

2 – 4 + 8 – 16 + …to 10 terms

Here, first term (a) = 2
t2
Common ratio (r) = t1 = –4 = – 2
2

No. of terms (n) = 10

Now, sum (Sn) = a(rn – 1)
r–1
2{(– 2)10 – 1}
= – 2–1

= 2(1024 – 1)
–3

= 2 × 1023
–3

= – 682.

(b) Here, the given series is
1 1
9 + 3 + 1 + …+ 9
Here, 1 1
first term (a) = 9

common ratio (r) = t2 = 3 × 1 × 9 = 3
last terms (l) = 9 t1 1 3
9

Now, sum (Sn) = lr – a 1
= r– 1 9
9× 3 –
1
3 –

= 27 × 9 – 1
9×2

= 243 – 1
18

= 242
18

= 121
9

74 Infinity Optional Mathematics Book - 10

2. Find the number of terms and the common ratio of the G.P. whose
first term is 7, last term is 189 and the sum of the series is 280.

Solution:

Here, first term (a) = 7

last term (tn) = l = 189

sum (Sn) = 280
lr – a
Now, sum (Sn) = r–1

or, 280 = 189r – 7
r– 1

or, 280r – 280 = 189r – 7

or, 91r = 273

∴ r=3

Again sum (Sn) = a(rn – 1)
3–1
7(3n – 1)
or, 280 =
or, 560 + 3–1
1 = 3n
7
or, 81 = 3n

or, 34 = 3n

∴ n = 4.
Hence common ratio (r) = 3 and no. of terms (n) = 4

3. The sum of first 6 terms of a GP is 126 and the sum of first 3 terms is
14, find the sum of first 12 terms.

Solution:

Here, sum of first 6 terms (S6) = 126

or, a(r6 – 1) = 126…(i)
r–1

sum of first 3 terms (S3) = 14

or, a(r3 – 1) = 14 …(ii)
r–1

Now, dividing equation (i) by (ii), we get
a(r6 – 1)
126
a(rr3––11) = 14

r–1

or, (r3 – 1)(r3 + 1) =9
r3 – 1

or, r3 + 1 = 9

Infinity Optional Mathematics Book - 10 75

or, r3 = 8

∴ r=2

Putting the value of r in equation (ii), we get

14 = a(23 – 1)
2–1

or, 14 = a × 7

∴ a=2

Now, sum of the first 12 terms

S12 = a(rn – 1)
r–1
2(212 – 1)
= 2– 1

= 2 × (4096 – 1)

= 2 × 4095

= 8190

4. The product of three numbers in a geometric series is 64 and their
sum is 14. Find the three numbers.

Solution:

Let, the three numbers in a G.P. be ar, a and ar.
By first case,

a × a × ar = 64
r
or, a3 = 64

∴ a=4

By second case,

ar + a + ar = 14

or, 4 + 4 + 4r = 14
r

or, 4 + 4r + 4r2 = 14r

or, 4r2 + 4r – 14r + 4 = 0

or, 2r2 – 5r + 2 = 0

or, 2r2 – 4r – r + 2 = 0

or, 2r(r – 2) – 1 (r – 2) = 0

or, (r – 2) (2r – 1) = 0

76 Infinity Optional Mathematics Book - 10

Either, r – 2 = 0

∴ r = 2

or, 2r – 1 = 0

∴ r = 1
2

When r = 2 and a = 4, then the three numbers are:

a = 4 = 2
r 2

∴ a = 4

∴ ar = 4 × 2 = 8

When r = 1 and a = 4, then the three numbers are:
2

ar = 4 = 8
2

∴ a = 4

∴ ar = 4 × 1 = 2
2

Hence, the three numbers in G.P. are 2, 4, 8 or 8, 4, 2.

Exercise 1.11

Section 'A'

1. Find the sum of the following series: 1 1 1
8 4 2
(a) 1 + 3 + 9 + … to 10 terms (b) + + + … to 15 terms

(c) – 4 + 8 – 16 + … to 8 terms (d) 128 + 64 + 32+ … to 10 terms
(e) 2 + 2 + 2 2 + … + 8 1
(g) 1 + 2 + 4 + … + 512 (f) 81 + 27 + 9 + … + 3

2. Find the value of: (b) 6 (c) 15 1n
∑ (– 3)n 2
(a) 4 ∑
∑ 2n n=2
n=1
n=1

3. (a) The first term and common ratio of the geometric series are 2 and 3
respectively. Find the sum of first 5 terms of the series.

(b) The first term, common ratio and the last term of the geometric

Infinity Optional Mathematics Book - 10 77

progression are 3, – 3 and – 729 respectively . Find the sum of the series.

4. (a) The common ratio and the sum of the first 7 terms of the geometric

progression are – 3 and – 1641 respectively. Find the first term.

(b) The common ratio and the sum of the first 10 terms of the geometric

progression are 1 and 1023 respectively. Find the first term.
2 1024

(c) Find the first term of the geometric series whose common ratio is 3, last

term is 189 and the sum of the series is 280.

(d) TFatT4en89hihrn12deemd11ffiitir5rrsh9es5sest1t51p6tfteieerrcarrestmnmtsivdpt,,eeeltllacrahytmss.eittvFsotetiueflenyrmtrdm.hmFeto,hifaagnetnnedhcdodeotmhtsmshueeemetmrcriosoiecounmsfsmirmetsahroot31iefine61osgt.rhewaoethmigooee.stoermiccoesmtreimrcieossnerariraeesti8oa1ri,es8121911,, last
5. (a) and
(b) 144

(c) If the sum of the first three terms is 7 and the first terms 1, find the

common ratio.

6. (a) How many terms of the series 32 + 48 + 72 + … will add up to 665 ?

(b) How many terms of the series 3 – 6 + 12 … must be taken so that the sum

may be – 63?

(c) In a GS the first term, common ratio and the sum are 1, 2 and 63

respectively. Find the number of terms in the series.

(d) In a G.P. the common ratio is 3 and the last term is 189. Find the number

of terms whose sum of the terms is 280.

7. (a) The first term, common ratio and the sum of the terms of geometric
1
progression are 64, 2 and127 respectively. Find the last term.

(b) Find the last term of the geometric series whose first term is 3, common

ratio is 2 and the sum of the series is 765.

Section 'B'

8. (a) In a G.P., 3rd term and 6th term are 12 and 96 respectively. Find the

sum of first 8 terms.

(b) Find the sum of first 10 terms of a GP where its 3rd term and 7th term are
1
4 and 4 respectively.

78 Infinity Optional Mathematics Book - 10

9. (a) In a G.P. the sum of first two terms is 6 and the sum of first four terms

is 30. Find the sum of first 7 terms of the progression.

(b) In a G.P. the sum of first three terms is 47 and the sum of first six terms
of first 10 terms of the series.
is 63 . Find the sum
32

10. (a) In a geometric series the sum of first 8 terms is 17 times the sum of first

4 terms. Find the common ratio.

(b) The sum of first 3 terms of a G.P. is 1 times the sum of first 6 terms. Find
9

the common ratio.

11. (a) In a GP, the sum of the three numbers is 7 and their product is 8. Find

the numbers.

(b) In a GP, the product of the three numbers is 1000 and their sum is 35.

Find the numbers.

(c) The product of three numbers in a geometric progression is 512 and the

sum of the first two number is 10. Find the three numbers.

12. (a) In a G.P, 6th term is 16 times the 2nd term and the sum of first seven
terms is 1427. Find the positive common ratio and the first term of the
series.

(b) The 7th term of a geometric series is 27 times the 4th term and the sum of
the first two terms is 8. Find the series.

(c) There are 8 terms in a geometric series altogether. The sum of the first
2 term is 6 and the sum of the last two terms is 384. Find the common

ratio.
(d) The sum of first four terms of a G.P. is 30 and that of the last four terms

is 960. If first term and last term are 2 and 512 respectively, find the
common ratio.

(e) There are 15 varieties of butterflies in a zoo. The number of each variety
being double the number of another veritey. If the number of the first
variety is 3, find the numbers in the last variety and also the total
number of all the varieties of butterflies in the zoo.

Infinity Optional Mathematics Book - 10 79

Relation between Arithmetic Mean (A.M.) and Geometric

Mean (G.M.)

Let us consider two numbers a and b. Arithmetic mean of a and b is A.M. = a + b .
2

Geometric mean (G.M.) = ab

Now,

A.M. – G.M. = a + b – ab
2
a + b – 2 ab
= 2

= (a)2 – 2 a . b+ (b)2
2
= 12( a– b)2

Since the square of any quantity is greater than or equal to zero.

A.M. – G.M. = 12( a – b)2 ≥ 0

or, A.M. – G.M. ≥ 0

∴ A.M. ≥ G.M.

When the two numbers a and b are equal

A.M. = G.M.

when a > b or b > a. then

A.M. ≥ G.M.

Hence, A.M. is always greater than or equal to G.M. between two positive real
numbers.

80 Infinity Optional Mathematics Book - 10

WORKED OUT EXAMPLES

1. Find the two numbers whose A.M. is 15 and G.M. is 9.

Solution: Alternate method
Let, the two numbers be a and b.

Then A.M. of a and b = a + b Let, the two numbers be a and b.
2
Then A.M. of a and b = a + b
or, 2
or, a + b = 30 15 = a + b
2
or, a = 30 – b …(i)
or, a + b = 30 …(i)
G.M. of a and b = ab
G.M of a and b = ab
or, 9 = ab
or, 9 = ab

or, ab = 81…(ii) or, ab = 81 …(ii)

Putting the value of a in equation (ii), Now, (a – b)2 = (a + b)2 – 4ab and
we get, or, (a – b)2 = (30)2 – 4 × 81
or, (a – b)2 = 900 – 324
(30 – b) b = 81 or, ( a – b)2 = 576
or, 30b – b2 = 81
or, b2 – 30b + 81 = 0 or, a – b = 576
or, b2 – 27b – 3b + 81 = 0
or, a – b = 24…(iii)
or, (b – 27) (b – 3) = 0
Now adding equation (i)
equation(iii) we get

Either b – 27 = 0 or, b – 3 = 0 a + b + a – b = 54
2a = 54
∴ b = 27 ∴ b=3 ∴ a = 27
Putting value of a in equation (i),
When b = 27, then we get ,

a = 30 – b = 30 – 27 = 3

When b = 3, then a + b = 30

a = 30 – b = 30 – 3 = 27 27 + b = 30

Hence, the two numbers are 27 and 3 ∴ b = 3
Hence, the two numbers are 27
or 3 and 27.
and 3.

2. If p, q, r are in A.P. and l, m, n are in G.P. then show that l q – r. mr – p.
np – q = 1.
p+r

Solution: when p, q, r are in A.P., then q = 2

or, p = 2q – r…(i)

Infinity Optional Mathematics Book - 10 81

When l, m, n are in GP, then'

m = ln
or, m = (ln)1/2
L.H.S. = lq – r × mr – p × np – q
= lq – r × (ln)12(q – 2q + r) × n2q – q – r
= lq – r . (ln)r – q . nq – r
= lq – r . lr – q . nr – q . nq – r

= lº. nº = 1 × 1 = 1 = R.H.S proved.

3. The sum of three numbers in on A.P. is 15. If 1, 4 and 19 are added to
each term respectively, then resulting numbers are in G.P. Find the
three numbers.

Solution:

Let, the three numbers in A.P. be

a – d, a and a + d

By first case,

a – d + a + a + d = 15

∴ a=5

Again by second case,

When 1, 4 and 19 are added to each term of A.P., then

a – d + 1, a + 4, a + d + 19 are in GP

or, 6 – d, 9, 24 + d are in GP

Then, 9 = (6 – d) (24 – d)

or, 81 = 144 + 6d – 24d – d2

or, d2 + 18d – 63 = 0

or, d2 + 21d – 3d – 63 = 0

or, (d + 21) (d – 3) = 0

Either d + 21 = 0 or, d – 3 = 0

∴ d = – 21 ∴d=3

When a = 5 and d = 3, then the three numbers are

a – d = 5 – 3 = 2

∴ a=5

∴ a+d=5+3=8

When a = 5 and d = – 21, then the numbers are

82 Infinity Optional Mathematics Book - 10

a – d = 5 – (– 21) = 26
∴ a=5
∴ a + d = 5 – 21 = – 16
Hence the three numbers in AP are 2, 5, 8 or 26, 5, –16.

4. An A.M. between two positive numbers p and q is twice their G.M.
Show that:

p : q = (2 + 3) : (2 – 3)

Solution:

A.M. of p and q = p+q
2
G.M of p and q = pq

By question, AM = 2 GM

or, p +2 q = 2 pq
or, p + q = 4 pq

or, (p + q)2 = 16pq

or, p2 + 2pq + q2 – 16pq = 0

or, p2 – 14pq + q2 = 0

or, p2 – 14 pq + 1 = 0 [Dividing both sides by q2]
q2

or, p 2 2 pq . 7 + 7² – 7² + 1 = 0
q
–

or, p –7 2
q
= 48

or, p – 7 = 48
q

or, p = 7 + 4 3
q

or, p = 4 + 4 3 + 3
q

or, p = (2)2 + 2 . 2 3 + ( 3 )2
q 4–3

or, p = (2 + ( 3 )2
q 22 – ( 3 )2

or, pq = (2 + ( 3)2 3 ))
(2 – ( 3)(2 + (

∴ p : q = (2 + 3) : (2 – 3) proved.

Infinity Optional Mathematics Book - 10 83

Exercise 1.12

1. (a) Find the two numbers whose arithmetic mean is 34 and geometric mean
is 16.

(b) If A.M. and G.M. between two numbers are 10 and 8 respectively, find
the two numbers.

(c) If the arithmetic mean and the geometric mean of the two numbers are
25 and 15 respectively then find the two numbers.

2. If a, b, c are in A.P. and x, y, z are in G.P. then show that xb – c yc – a za – b =1
3. The sum of three numbers in A.P. is 21. If the 2nd number is reduced by 1

and 3rd number increased by 1, then the three numbers become G.P. Find the
numbers.
4. The sum of three numbers in A.P. is 21. If 1, 2, and 15 are added to each term
respectively, the resulting numbers are in G.P. Find the three numbers.
5. The product of three numbers in G.P. is 216. When 1 and 2 are added to first
term and second term respectively, the resulting numbers are in A.P. Find the
three numbers.
6. The second, fourth and ninth terms of an AP are in G.P. Calculate the common
ratio of G.P.
7. The ratio of A.M. and G.M. of two numbers is 5:3 . Find the ratio of two
numbers
8. The two positive numbers are in the ratio of 1:9 and the sum of their A.M. and
G.M. is 96. Find the numbers.
9. The sum of the four consecutive terms of a G.P. is 30. If the A.M. of the first
term and the last term is 9, find the common ratio.
10. An A.M. of two positive numbers is more than their G.M. by 2. If the difference
between the two numbers is 12, then find them.
11. The arithmetic mean of two numbers is 50 and geometric mean is 80% of
arithmetic mean. Find the four arithmetic mean between these two numbers.
12. An A.M. between two positive numbers a and b is twice their G.M. Show that:
a:b = (2 + 3): (2 – 3)

84 Infinity Optional Mathematics Book - 10

1.4 Linear Inequation and Graph of Quadratic Equation

Introduction

We know that an equation is a statement involving the sign of equality (=). There
may be one or more than one variable in an equation. The value of the variables
which satisfies an equation are called the solutions of the equation. The solution
of an equation may or may not be unique. For example, x = 4 is a solution of the
equation 7x – 28 = 0. The equation x2 + y2 = 25 contains two variables. This has
infinitely many solutions and x = 3, y = 4 is one of the solutions.
A statement involving a sign of inequality either > or > or < or < is called an
inequation or an inequality. An inequality may contain one variable or more than
one variable. The values of the variables which satisfies an inequation are called
the solution of the inequation.

Intervals

If a, b are real numbers such that a < b, then the set {x : a< x < b} is called the open
interval from a to b and is written as (a, b). The points a and b are called the end
points of the open interval (a, b).
The set {x: a < x < b} is called the closed interval from a to b and is written as [a, b].
The points a and b are called end points of the closed interval [a, b].

ab ab
open interval: (a, b) close interval: (a, b)

Similarly, we define semi-open (or semi closed) intervals (a, b] and [a, b).
If a is any real number, then the sets of the type {x : x < a}, {x : x < a}, {x: x > a} and
{x: x > a} are called infinite intervals and are respectively written as [– ∞, a), (–∞,
a], (a, ∞), [a, ∞). The set R of real numbers is also an infinite interval, written as
(–∞, ∞).

Linear Inequation in One Variable

A linear inequation in one variable (x) is of the form either ax + b > 0 or ax + b > 0
or ax + b < 0, ax + b < 0. A linear inequation in one variable is solved by collecting
all terms containing the variable on the left side and the constants on the right side.
For example, 2x + 5 > 0, 2x – 5 < 6, 7x + 2 < 0, 3x + 5 > 0 etc. The graph of a linear
inequality is half ray in the real number line. Let us show the following inequaitons

Infinity Optional Mathematics Book - 10 85

in one variable with their graphs.
(a) Graph of 2x – 6 > 0
Here, 2x – 6 > 0
2x > 6
x > 3
Here, all the values of x greater than 3 satisfy the inequality. But the point x

= 3 is not included i.e. x = {4, 5, 6, ....}

–5 –4 –3 –2 –1 0 1 2 3 4 5 6 7

(b) Graph of 2x + 8 < 2

Here, 2x + 8 < 2

or, 2x < 2 – 8

or, 2x < – 6

or, x < – 6
2

∴ x < – 3
Here, all the values of x less than – 3 satisfy the inequality. But the point x

= – 3 is not included.
i.e. x = {–4, –5, –6, .....}

–7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5
The two graphs (a) and (b) are called open half line.
(c) Graph of 3x > 9
Here, 3x > 9

or, x > 9
3

or, x > 3

∴ x = {3, 4, 5, .....]

–6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7
(d) Graph of 5x – 19 < 1
Here, 5x – 19 < 1
or, 5x < 1 + 19
or, 5x < 20

or, x < 20
5

or, x < 4

86 Infinity Optional Mathematics Book - 10

∴ x = {4, 3, 2 ......]

–5 –4 –3 –2 –1 0 1 2 3 4 5 6
The two graphs (c) and (d) are called closed half line.

Linear Inequations in Two Variables

The equation ax + by + c = 0 is called general form of the linear equation in x and
y. The inequation of the form ax + by + c > 0 or ax + by + c < 0 or ax + by + c > 0 or
ax + by + c < 0 is called linear inequation in two variables x and y where a, b and c
are real constant. Some examples of linear inequations are 3x + 2y < 5, 5x – y > 2,
6x + 7y < 3, x + y > 3 etc.

An ordered pair (p, q) is called solution of an inequality in x and y when the inequality
holds true when p and q are replaced for x and y respectively in the inequation. For
example, (2, 1) is a solution of the inequality 5x – y > 2 since 5 × 2 – 1 > 2.

i.e. 9 > 2 is true.

The solution set of a linear inequality ax + by + c > 0 or ax + by + c < 0 or ax + by +
c > 0 or ax + by + c < 0 is one of the two half planes divided by the line ax + by + c =
0, called the boundary line.

In the inequality having > or < sign, we take a dotted line or broken line and the
inequality having > or < sign, we take solid line.

To determine the solution of given linear inequallity in two variables
graphically, first we draw the boundary line. Then we choose any point
on the plane not lying on the boundary line as a testing point. Generally,
(0, 0) is taken as a testing point for the line that does not pass through origin and
(1, 0) is taken as a testing point that passes through the origin. If the inequality
is satisfied by this testing point, (i.e. true condition) then out of the two regions
(half plane) divided by the boundary line, the solution is the one which contains
the testing point. If the inequality is not satisfied, (i.e. false condition), then the
solution set is the other half plane not containing the testing point.

Infinity Optional Mathematics Book - 10 87

WORKED OUT EXAMPLES

1. Draw the graph of the following inequalities Y

(a) x > 5 X' O X
Solution: x>5
For x < 5, the boundary line is x = 5 which
X
gives a dotted vertical line parallel to
y-axis. Also, x = [6, 7, 8, ....] which shows y<5
X
the region to the right of this line. Y'
Y X
y>–4
(b) x ≤ –3 O

Solution: Y'
For x < – 3, the boundary line is x = –3, X' Y

which gives a solid vertical line parallel x < –3 O

to y-axis. Also, x = {–3, –4, – 5, ....} which

shows the region to the left of this line.


(c) y < 5 X'
Solution:
For y < 5, the boundary line is y = 5, which

gives a dotted horizontal line parallel to
x-axis. Also, y = {4, 3, 2, .......} which shows
the region towards the origin.


(d) y ≥ –4 X' Y'
Y
Solution:
For y > – 4, the boundary line is y = –4, O

which gives a solid horizontal line parallel
to x-axis. Also, y = {–4, –3, –3, ....], which
shows the region towards the origin.


Y'

88 Infinity Optional Mathematics Book - 10

(e) 3x + 4y < 12

Solution:
Here, 3x + 4y < 12
The boundary line of 3x + 4y < 12 is 3x + 4y = 12

or, y = 12 – 3x x 0 4 –4
4 y306

(x, y) (0, 3) (4, 0) (–4, 6)
Taking (0, 0) as a testing point, we get

3 × 0 + 4 × 0 < 12

0 < 12 (True)

Here, the testing point satisfies the given inequality, so the half plane

represented by 3x + 4y < 12 contains testing point (0, 0).
Y

3x + 4y < 12
X' O X

Y'

(f) 2x – 3y > 6

Solution: Here, 2x – 3y > 6
The boundary line of 2x – 3y < 6 is 2x – 3y = 6,

or, y = 2x – 6 x 0 3 –3
3 y –2 0 –4
(x, y) (0, –2) (3, 0) (–3, –4)

Taking (0, 0) as a testing point, we get

2 × 0 – 3 × 0 > 6

0 > 6 (False)

Here, the testing point does not satisfy the given inequality. So, the half plane

Infinity Optional Mathematics Book - 10 89

represented by 2x – 3y > 6 does not contain testing point (0, 0).
Y

2x – 3y > 6
X' O X

Y'

(g) 5x + 2y ≥ 0

Solution:

Here, 5x + 2y > 0

The boundary line of 5x + 2y > 0 is 5x + 2y = 0
– 5x
or, y = 2 x 0 2 –2
0 –5 5
Taking (1, 0) as testing point, we get y (0, 0) (2, –5) (–2, 5)
5×1+2×0>0
(x, y)

or, 5 > 0 (True)

Y

X' O X

Y'
5x + 2y > 0

90 Infinity Optional Mathematics Book - 10

System of linear inequalities Y X
O
In the adjoining figure, the
shaded region is formed by three Y'
inequalities. Hence two or more
than two inequalities of the same
variables having a common solution
set is called the system of linear X'
inequalities. In this system, the
plane regions determined by the set
of the inequalities are drawn on the
same graph. If the two inequalities
intersect, the point of intersection of
their boundary lines is called their vertex.

WORKED OUT EXAMPLES

1. Draw the graph of the following inequalities and show the common
solution set.

2x + 3y < 6, 2x – y > 2.
Solution:

The boundary line of 2x + 3y < 6 is 2x + 3y = 6

or, y = 6 – 2x
3
x 0 3 –3
2 0 4
Taking (0, 0) as testing point, we get y (0, 2) (3, 0) (–3, 4)

2×0+3×0<6 (x, y)

0 < 6 (True)

The boundary line of 2x – y > 2 is 2x – y = 2

or, y = 2x – 2 x01 2

Taking (0, 0) as testing point, we get y –2 0 2
2 × 0 – 0 > 2 (x, y) (0, –2) (1, 0) (2, 2)

0 > 2 (False)

Infinity Optional Mathematics Book - 10 91

Y

X' O X

Y'
The shaded region represents the solution set.

2. Draw the graph of x + y < 6, x – y > – 2, x > 0, y > 0.

Solution: The boundary line of x + y < 6 is x 0 6 2
x+y=6 y 6 0 4
or, y = 6 – x (x, y) (0, 6) (0, 6) (2, 4)
Taking (0, 0) as testing point, we get
–2 –4
0+0<6 x 2 0 –2
or, 0 < 6 (True) y 4 (–2, 0) (–4, –2)
The boundary line of x – y > – 2 is x – y = –2 (x, y) (2, 4)
or, y = x + 2

Taking (0, 0) as testing point, we get

0 – 0 > – 2 0 > – 2 (True)

The boundary line of x > 0 is x = 0 which is the equation of y-axis and show

that region to the right of y-axis.

The boundary line of y > 0 is y = 0 which is the equation of x -axis and show

the region above x-axis.

Y

X' O X

Y'
The shaded region represents the solution set.

92 Infinity Optional Mathematics Book - 10

Exercise 1.13

Section 'A'

1. Draw the graph of each of the following inequalities.

(a) x < 0 (b) y > 0 (c) x < 3 (d) x > –4

(e) y > 0 (f) 3 – 5y > 5 – 3y (g) 3x < – 6 (h) –4 < y < 3

2. Draw the graph of the following inequalities

(a) 3x – 2y > 6 (b) x – y < 5 (c) 2x + y < 6

(d) 2x + 5y + 10 < 0 (e) 3x – y < 0 (f) 2x + 5y > 0

Section 'B'

3. Draw the graphs of the following inequalities and shade the

common solution set.

(a) x + y > 6 and y > – 1 (b) x – y > 1 and x + y < 1

(c) 2x – y < 0 and 2x + 3y < 6 (d) 2x + 3y < 12, x > 0, y > 0

(e) x + y < 5, x < 4, y > 1, x > 0 (f) 3x + 4y < 24, 0 < y < 6, 0 < x < 6

Linear Programming

When the investor starts any business activities, they want to make maximum
profit with minimum investment, maximum production with a use of limited
resources. Such type of problems can be solved by the mathematical technique
called linear programming. This technique was started after the second world
war by the Russian mathematician L.V. Kantorovich. Later it was further
developed by the American mathematician George B. Dantzig. Some important
terms used in linear programming are :
(i) Objective function
The linear function F= ax + by + c which is to be maximized or minimized

is called the objective function.
(ii) Constrains: The given inequalities are called constrains.
(iii) Solution: The values of x and y which satisfy the inequalities (constraints)

are called the solution. The solutions are the vertices of the convex polygonal
region.
(iv) Feasible solution: A set of values of x and y which satisfy the set of
constraints or conditions is called a feasible solution.
(v) Optimal solution: Any feasible solution, which optimizes (maximizes or
minimizes) the objective function is called on optimal solution.

Infinity Optional Mathematics Book - 10 93

Convex Region:
A plane region is said to be convex if it contains every line segment joining any two
points in it.
Example: Rectangular, Triangular and Circular regions are convex.


but a star-shaped region is not convex.


WORKED OUT EXAMPLES

1. Calculate the maximum and minimum value of the objective function
F = 3x + 5y – 2 subject to x – y < 2, x + y < 4, x > 0, y > 0.

Solution: x 2 4 1
The boundary line of x – y < 2 is x – y = 2 y 0 2 –1
(x, y) (2, 0) (4, 2) (1, –1)
or, y = x – 2 .......... (i)

Taking (0, 0) as testing point, we get

0 – 0 < 2

0 < 2 (True)

So, the half plane of the given inequality contains origin.

The boundary line of x + y < 4 is x + y = 4

or, y = 4 – x .......... (ii) x 4 2 –1
Taking (0, 0) as testing point, we get y 0 2 5
0+0<4 (x, y) (4, 0) (2, 2) (–1, 5)

0 < 4 (True)

So, the half plane of the given inequality contains origin.

For x > 0

∴ x = {0, 1, 2, ......} ........ (iii)

For y > 0

∴ y = {0, 1, 2, ......} ......... (iv)

94 Infinity Optional Mathematics Book - 10

(ii) Y (i)
C

B

X' OA y>0 X

x>0

Y'
The shaded region OABC is the required feasible region (solution set) whose

vertices are O(0, 0), A(2, 0), B(3, 1) and C(0, 4)



Vertices Objective function Remark

F = 3x + 5y – 2

O(0, 0) 3×0+5×0–2=–2 Minimum
A(2, 0) 3×2+5×0–2=4 Maximum
B(3, 1) 3 × 3 + 5 × 1 – 2 = 12
C(0, 4) 3 × 0 + 5 × 4 – 2 = 18

Hence, the maximum value is 18 at C(0, 4) and minimum value is –2 at

O(0, 0). Y
2. In the given diagram OABC

is a feasible region whose B(2, 5)
coordinates of B and C are
(2, 5) and (0, 3) respectively. If C(0, 3)

one of the inequalities is x + y < 7,

then find the other inequalities X' O AX
and hence maximize P = 3x – y.

Solution:

Here, the given inequality is

x + y < 7. Its boundary line is x + Y'
7 = 7.

Infinity Optional Mathematics Book - 10 95

or, 7x + y = 7
7 7

or, x + y = 1 x y
7 7 a b
Comparing with + = 1, we get

x-intercept (a) = 7 and y-intercept (b) = 7.

In the figure, the line AB has x -intercept 7 and y-intercept 7. So, the equation
of the line AB is x + y < 7.

The equation of the line BC joining B(2, 5) and C(0, 3) is

y – y1 = y2 – y1 (x – x1)
x2 – x1
or,
y –5 = 3– 5 (x – 2)
or, 0– 2
or, –2
or, y – 5 = –2 (x – 2)
∴
y–5=x–2

x–y+3=0

Equation of the line BC is x – y + 3 = 0.

Taking (0, 0) as testing point, we get

x – y + 3 0 – 0 + 3 = 3 > 0

∴ x – y + 3 > 0 is the inequation of line BC.

Here the shaded region lies in first quadrant and the equation of x-axis and
y-axis are y = 0 and x = 0 respectively. So, their inequalities are y > 0 and
x > 0.

Hence, the inequalities are

x + y < 7, x – y + 3 > 0, x > and y > 0.

From the graph, OABC is a required solution set whose vertices are O(0, 0),
A(7, 0), B(2, 5) and C(0, 3).



Vertices Objective function Remark

O(0, 0) F = 3x – y Maximum
A(7, 0) Minimum
B(2, 5) 3×0–0=0
C(0, 3) 3 × 7 – 0 = 21
3×2–5=1
3 × 0 – 3 = –3

96 Infinity Optional Mathematics Book - 10

Exercise 1.14

1. Determine the maximum and minimum values of the objective
function F = 2x + 3y.

Y (b) Y
(a)
D(3,6)

C(0,3) C A(1,4)

B(3,2) C(5,3)

X' O A(4,0) X X' O B(3,1) X

Y' Y'

(c) Y (d) Y

B(6,6)

C(4,3) C(6,4)
X' O D(3,3)

A(6,1)

X X' O A(6,0) B(10,0) X

Y' Y'

2. Maximize and minimize the following objective functions under the
given constraints.
(a) P = 5x – 2y subject to 2x – 3y < 6, y < 2, x > 0, y > 0
(b) P = 6x + 2y subject to x + 2y > 5, y < 2x, x < 5
(c) P = 9x + 7y subject to x + 2y < 7, x – y ≤ 4, x > 0, y > 0
(d) P = 2x + y subject to x + y > 6, x – y > 4, x < 6
(e) P = x + y + 2 subject to x + y < 4, x > 0, y > 0.

3. (a) OABC is a feasible region. Write down the inequalities to represent the
given graph,

Also find the maximum value of p = 9x + 3y.

Infinity Optional Mathematics Book - 10 97

Y

(0,4)

(5,1)

X' O A(4,0) X

Y'

(b) From the given graph, find the inequalities which represents the shaded

region and calculate the maximum and minimum value of P = 4x + 9y.

Y

C
B

X' 0A X

Y'

Quadratic Equation and Graphs:

Quadratic Equation

The equation of the form ax2 + bx + c = 0 (where a ≠ 0) is called the quadratic

equation of second degree in one variable. We always get two values of the variables

when we solve quadratic equation. The values of the variable are also called roots

of quadratic equation.

Graph of quadratic function
The equation y = ax2 + bx + c is called the general from of a quadratic function.
Hence, a, b and c are constants. The nature of the graph of quadratic function is
called parabola.

98 Infinity Optional Mathematics Book - 10

There are three forms of quadratic functions

(a) y = ax2 (b) y = ax2 + c (c) y = ax2 + bx + c

(i) Graph of y = ax2

y = x2 where a = 1,

x 0 1 –1 2 –2

y01144

(x, y) (0, 0) (1, 1) (–1, 1) (2, 4) (–2, 4)



Y

y = x2

X' O X

(ii) y = –x2 where a = –1 Y'

x0 1 –1 2 –2
–1 –1 –4 –4
y0 (1, –1) (–1, –1) (2, –4) (–2, –4)

(x, y) (0, 0)

Y

X' O X

y = –x2

Y'
From the above two graphs, the following informations are found.
(i) The nature of the graph of y = ax2 is a parabola.
(ii) The turning point (vertex) of parabola is at origin.

Infinity Optional Mathematics Book - 10 99

(iii) When a is +ve, the parabola is upward. –2
(iv) When a is –ve, the parabola is downward. 11
(v) The parabola is symmetrical about y-axis. (–2, 11)
(b) Graph of y = ax2 + c
(i) y = 2x2 + 3. Where a = 2 and c = 3.

x 0 1 –1 2
y 3 5 11 7
(x, y) (0, 3) (1, 5) (–1, 11) (2, 7)

Y

y = 2x2 + 3

X' O X

Y'
Similarly, draw the graph of the following equations.
(i) y = 3x2 – 5
(ii) y = –x2 + 2
(iii) y = –2x2 – 1
Also, fill in the blanks from the graph of the above equations.

Figure no. Vertex of parabola Upward or downward Line of symmetry

(i)
(ii)
(iii)

100 Infinity Optional Mathematics Book - 10