## NCERT Solutions for Class 11 Maths Chapter 9 Sequence and Series Exercise 9.3

Chapter 9 Sequence and Series Exercise 9.3 NCERT Solutions for Class 11 Maths is very useful for completing your hometask as these Class 11 Maths NCERT Solutions prepared by Studyrankers experts are detailed and accurate. NCERT Solutions are best way through which you can look around the chapter easily and cover all the important topics.

1. Find the 20

Tn = ar

T

T

1. Find the 20

^{th}and n^{ th}terms of the G.P. 5/2, 5/4, 5/8, ...........**Answer**Tn = ar

^{n-1}, we have, a = 5/2, r = 5/4÷ 5/2 = 1/2;T

_{20}= 5/2(1/2)^{20 – 1}= 5/2 . 1/2^{19}= 5/2^{20}T

_{n}= 5/2 (1/2)^{n -1}= 5/2 . 1/2^{n-1 }= 5/2^{n}2. Find the 12

^{th}term of a G..P. whose 8

^{th}term is 192 and the common ratio is 2.

**Answer**

Let a be the first term and r be the common ratio of a G.P.

Then T

_{8}= ar

^{8–1 }= ar

^{7}= 192 … (i)

Now T

_{12}= ar

^{12 – 1 }= ar

^{11}= (ar

^{7})r

^{4}

= 192 × 2

^{4}[ ∵ r = 2]

= 192 × 16 = 3072

3. The 5

^{th}, 8

^{th}and 11

^{th}terms of a G.P. are p, q and s, respectively. Show that q

^{ 2}= ps.

**Answer**

Let a be the first term and r the common ratio of G.P. then

T

_{5}= p => ar

^{4}= p … (i)

T

_{8}= q => ar

^{7}= q … (ii)

T

_{11}= s => ar

^{10}= s … (iii)

Now, q

^{2}= (ar

^{7})

^{2}[Using (ii)]

=> q

^{2}= a

^{2}r

^{14}

=> q

^{2}= (ar

^{4})(ar

^{10})

=> q

^{2}= ps [Using (i) and (iii)]

4. The 4

^{th}term of a G.P. is square of its second term, and the first term is –3. Determine its 7

^{ th}term.

**Answer**

Let a be the first term and r the common ratio of G.P. then

T

_{4}of G.P. = (T

_{2}of G.P.)

^{2}and a = – 3

ar

^{3}= (ar)

^{ 2}

=> ar

^{3}= a

^{2}r

^{2}

=> r = a = -3

Now, T

_{7}= ar

^{6}= (–3)(–3)

^{6}= (–3)

^{7}= –2187.

5. Which term of the following sequences:

(a) 2, 2√2, 4.... is 128?

(b) √3, 3, 3√3,.... is 729?

(c) 1/3, 1/9, 1/27 …….. is 1/19683?

**Answer**

(a) The G.P, is 2, 2√2, 4, .....

The first term = a = 2,

The common ration r = √2

n th term = ar

^{n – 1 }= 128

or 2.(√2)

^{n -1}= 128 or 2

^{n-1/2}= 64 = 2

^{6}

=> (n – 1)/2 = 6 or n -1 = 12 or n = 13

(b) (b) Here a= √3, r = √3. Let T

_{n}= 729

=> ar

^{n-1 }= 729

=> (√3)(√3)

^{n-1}= 729

=> (√3)

^{n}= (9)

^{3}

=> 3

^{n/2 }= (3

^{2})

^{3}= 3

^{6}

=> n/2 = 6 => n = 12

Hence, the 12

^{th}term is 729

(c) (c) a = 1/3, r = (1/9)/(1/3) = 1/9 × 3/1 = 1/3

let T

_{n}= 1/19683 => ar

^{n-1 }= 1/19683

=> 1/3(1/3)

^{n-1}= 1/19683

=> (1/3)

^{n}= (1/3)

^{9}=> n = 9

6. For what values of x, the numbers –(2/7), x, -(2/7) are in G.P?

**Answer**

The numbers -2/7, x, -7/2 will be in GP

If x/-(2/7) = -7/(2/x) => x2 = -(7/2) × –(2/7) = 1 => x = ±1

7. Find the sum to indicated number of terms in each of the geometric progressions is 0.15, 0.015, 0.0015, .......... 20 terms.

**Answer**

We have a = 0.15, r = .015/.15 = 0.1 < 1, n = 20

S

_{n}= (a(1 –r

^{n}))/(1 – r), r < 1;

S

_{20}= (0.15[1 – (0.1)

^{20}])/1 = 0.1 = (0.15[1 – (0.1)

^{20}])/0.9

= 1/6 [1 – (0.1)

^{20}]

8. Find the sum to indicated number of terms in the following geometric progression:

√7, √21, 3√7,....... n terms.

**Answer**

Here a = √7

R = √21/√7 = √(21/7) = √3 > 1, n = n

S

_{n}= (a(r

^{n}– 1))/(r – 1) = (√7[(√3)

^{n}– 1])/(√3 – 1)

= (√7(√3 + 1)[(√3)

^{n}– 1])/((√3 – 1)(√3 + 1))

= (√7(√3 + 1)[(√3)

^{n}– 1])/2

9. Find the sum to indicated number of terms in the following geometric progression: 1, – a, a

^{2}, – a

^{3}, ....... n terms (if a ≠ – 1)

**Answer**

G.P. is 1, – a, a

^{2}, – a

^{ 3}, .......

Now first term A = 1, r = – a

S

_{n}= (a(1 – r

^{n}))/(1 –r)

= (1[1-(-a)

^{n}]/(1-(-a)) = (1-(-a)

^{n})/(1 + a)

10. Find the sum to indicated number of terms in the following geometric progression x

^{ 3}, x

^{ 5}, x

^{ 7}, ..... n terms (if x ≠ ± 1).

**Answer**

Here a = x

^{3}, r= x

^{5}/x

^{3}= x

^{2}, x ≠ ± 1

S

_{n}= (a(1 – r

^{n}))/(1 – r)

S

_{n}= (x

^{3}(1 – x

^{2n})/(1 – x

^{2})

11.

**Answer**

12. The sum of first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms.

**Answer**

Let the first three terms of a G.P. be a/r, a, ar.

Product of three terms = a/r × a × ar = 1

a

^{3}= 1 => a = 1

Sum of these term1/r + 1 + r = 39/10 (put a = 1)

Multiplying by 10r; 10 + 10r + 10r

^{2}= 39r

=> 10r

^{2}+ 10r – 39r + 10 = 0

=> 10r

^{2}– 29r + 10 = 0

=> (2r – 5)(5r – 2) = 0 => r = 5/2 or 2/5

When r = 5/2 the term of G.P. are 2/5, 1, 5/2

When r = 2/5, the terms of G.P. are 5/2, 1, 2/5.

13. How many terms of G.P. 3, 3

^{2}, 3

^{3}, .......... are needed to give the sum 120?

**Answer**

Let n be the number of terms of the G.P. 3, 3

^{2}, 3

^{ 3}, .......... makes the sum = 120

We have a =3, r = 3

S = (a(r

^{n}-1))/(r – 1), r > 1;

Sum = (3(3

^{n}– 1))/(3 – 1) = 120

Or 3/2 (3

^{n}– 1) = 120

Multiplying both sides by 3/2

∴ 3

^{n}– 1 = 80

∴ 3

^{n}= 80 + 1= 81 = 3

^{4}=> n = 4

∴ Required number of terms of given G. P. is 4.

14. The sum of the first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.

**Answer**

Let a be the first term and r be the common ratio.

S

_{3}= 16

(a(1 – r

^{3}))/(1- r) = 16 …….(i)

S

_{6}– S

_{3}= 128 => (a(1 – r

^{6}))/ (1 – r) – 16 = 128,

i.e., (a(1 –r

^{6}))/(1 – r) = 144 ….(ii)

(ii) ÷ (i) gives (1 – r

^{6})/(1 – r

^{3}) = 144/16 = 9/1

=> ((1 – r

^{3})(1 + r

^{3}))/(1 – r

^{3}) = 9/1 => 1 + r

^{3}= 9

=> r

^{3}= 8 => r

^{3}= 2

^{3}=> r = 2

Thus, common ratio = 2

As, r = 2, S

_{3}= (a(r

^{3}– 1))/(r – 1) = 16

=> (a(2

^{3}– 1)/(2 – 1) = 16 => (a(8 – 1))/1 = 16

=> a(7) = 16 => a = 16/7

S

_{n}= (a(r

^{n}– 1))/r – 1 = 16/7 . (2

^{n}– 1)/(2 – 1) = 16/7 .(2

^{n}– 1).

15. Given a G.P. with a = 729 and 7

^{th}term 64, determine S

_{7}

**Answer**

a = 729, T

_{7}= 64.Let common ratio = r

=> ar

^{6}= T

_{7}= 64 => 729 r

^{ 6}= 64 => r

^{ 6}= 64/729

=> r

^{6}= (2/3)

^{6}=> r = 2/3 < 1

S

_{n}= (a(1 – r

^{n}))/(1 – r)

S

_{7}= (729[1 – (2/3)

^{7}])/1 – 2/3

= 729 × (3(37– 2

^{7}))/3

^{7}

= (3

^{6}.3(3

^{7}– 3

^{7})/3

^{7}= 3

^{7}– 2

^{7}= 2187 – 128 = 2059.

16. Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.

**Answer**

Let a be the first term and r be the common ratio.

Also, S2 = –4, T

_{5}= 4T

_{3}

=> (a(1 – r

^{2}))/(1 – r) = -4 => a(1 + r) = -4

=> a( 1 – r) = -4, ar

^{4}= 4ar

^{2}=> r

^{2}= 4

=> r = 2

Now, when r = 2 : a (1 + 2) = -4 => a = -4/3

Thus, the G.P. is -4/3 , -8/3, -16/3 ,………..

When r = -2 : a(1 – 2) = -4 => a(-1) = -4

=> a = 4

Thus, the G.P. is 4, -8, 16, -32, 64, ……….

17. If the 4

^{th}, 10

^{th}and 16

^{th}terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.

**Answer**

Let a be the first term and r be the common ratio.

T

_{4}= x => ar

^{3}= x … (i)

T

_{10}= y => ar

^{9}= y … (ii)

T

_{16}= z = ar

^{15}= z …(iii)

Now, x, y, z will be in G.P.

If ar

^{3}, ar

^{9}, ar

^{15}are in G.P.

i.e., ar

^{9}/ar

^{3}= ar

^{15}/ar

^{9}=> r

^{6}= r

^{6}, which is true.

18. Find the sum of n terms of the sequence 8, 88, 888, 8888, .........

**Answer**

Let S be the sum of n-terms of the series,

8 + 88 + 888 + 8888 + .......

S = 8 + 88 + 888 + 8888 + ............... to n terms

= 8(1 + 11 + 111 + 1111 + ....... to n terms)

= 8/9 (9 + 99 + 999 + 9999 + ......... to n terms)

= 8/9 [(10 – 1) + (10

^{2}– 1) + (10

^{3}– 1) + …………… to n terms]

= 8/9 [(10 + 10

^{2}+ 10

^{3}+ …………….. to n terms ) – (1 + 1+ 1 ……….. to n terms)]

= 8/9 [(10(10

^{n }– 1))/(10 – 1) - n]

S = 8/9 [10/9 (10

^{n}– 1) – n].

19. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, 1/2.

**Answer**

Sum of product of corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, 1/2 is

S = 2.128 + 4.32 + 8.8 + 16.2 + 32 where S denotes the sum of these term.

S = 256 + 128 + 64 + 32 + 16 =

It is a G.P

a = 256, r = 1/2,

S = (a(1 –r

^{5}))/(1- r) = 256 × 2 × (1 – 1/32)

= 256 × 2 × 31/32 = 16 × 31 = 496.

20. Show that the products of the corresponding terms of the sequences a, ar, ar

^{2}.......... ar

^{n – 1 }and A, AR, AR

^{2}, ...... AR

^{n – 1 }form a G. P, and find the common ratio.

**Answer**

The two sequences are a, ar, ar

^{2}........ ar

^{n – 1 }and A, AR, AR

^{2}, ..........AR

^{n – 1}

∴ Sum of the products of the corresponding term of these sequence

aA + aA(rR)+ aA(r

^{2}R

^{2}) + .......+ a Ar

^{n– 1}R

^{n – 1}

= aA + aA (rR) + aA (rR)

^{ 2}+ ........ + aA(rR)

^{ n– 1}

This is a G.P. whose first term is aA and the common ratio = aA(rR) ÷ aA = rR

21. Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4

^{th}by 18.

**Answer**

Let a be the first term and r the common ratio

T

_{n}=ar

^{n –1}, T

_{2}= ar, T

_{3}= ar

^{2}and T

_{4}= ar

^{3}

Since third term is greater than the first by 9

T

_{3 }= T

_{1}+ 9 => ar

^{2}= a + 9 …(i)

Also the second term is greater than the fourth by 18

∴ T

_{2}= T

_{4}+ 18 => ar = ar

^{3}+ 18 …(ii)

Multiplying (i) by r, we get,

ar

^{3}= ar + 9r …,(iii)

from (ii) & (iii)

ar = ar + 9r + 18 => 0 = 9r + 18

=> r = -18/9 = -2

Putting r = -2 in (i)

a(-2)

^{2}= a + 9 => 4a = a + 9 => 3a = 9

=> a = 3

Now, T

_{2}=ar = 3(-2) = - 6;

T

_{3}= ar

^{2}= 3(-2)

^{2}= 12;

T

_{4}= ar

^{3}= 3(-2)

^{3}= -24

∴ The required terms are 3, -6, 12 and -24.

22. If the p

^{ th}, q

^{ th}and r

^{ th}terms of a G..P. are a, b, and c respectively. Prove that

a

^{q-r }b

^{r-p }c

^{p-q}= 1.

**Answer**

Let A be the first term and R the common ratio of G.P.

T

_{p}= a => AR

^{p – 1}= a …(i)

T

_{q}= b => AR

^{q – 1}= b …(ii)

T

_{r}= c => AR

^{r – 1}= c …(iii)

a

^{q – r}. b

^{r – p}. c

^{p – q}

= (AR

^{p – 1})

^{q – r}. (AR

^{q – 1})

^{r – p}. (AR

^{r – 1})

^{p – q}

= A

^{q – r + r – p + p – q}× R

^{pq – pr – q + r + qr – pq –r + p + pr – qr – p + q}

=A

^{0}R

^{0}= 1. 1 = 1.

23. If the first and the n th term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P

^{2}= (ab)

^{n}

**Answer**

Let a be the first term and r the common ratio of a G.P.

T

_{1}= a and T

_{n}= b

ar

^{n – 1 }= b … (i)

Also, P = T

_{1}. T

_{2}. T

_{3}. .............T

_{n}

= a. ar. ar

^{2}, ..........ar

^{n – 1}

= a

^{ n}. r

^{1 + 2 + 3 + .........+ (n –1)}

Since, 1 + 2 + 3 + .........+ (n – 1) = (n(n -1))/2

∴ P = a

^{n}. r

^{n(n – 1)/2}

∴ P

^{2}= a

^{2n}. r

^{n(n – 1) }= [a

^{2}r

^{n-1}]

^{n}

= [a. ar

^{n -1}]n = (a . b)

^{n}[using (i)]

Hence, P

^{2}= (ab)

^{n}

24. Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)

^{th}to (2n)

^{ th}term is 1/r

^{n}

**Answer**

Sum of the first n terms of a G.P. = a + ar + ar

^{2}+ ..............+ ar

^{n – 1 }= (a(1- rn))/(1 – r) ………(i)

T

_{n + 1 }= ar

^{n}, T

_{n + 2 }= ar

^{n +1}, ..........T

_{2n}= ar2

^{n – 1}

∴ Sum of terms from (n + 1)

^{th}to (2n)

^{ th}terms

=T

_{n + 1 }+ T

_{n + 2}+ T

_{n + 3 }+ ..........+ T

_{2n}

= ar

^{n }+ ar

^{n + 1 }+ ar

^{n + 2 }+ .........+ar

^{2n – 1}

= (ar

^{n}(1 – r

^{n}))/(1 – r)…(ii)

No. of terms = n

Dividing (i) by (ii),

(sum of first n terms of a G.P.)/(Sum of terms from (n + 1)

^{th}to (2n)

^{th}term)

=

25. If a, b, c and d are in G.P. show that

(a

^{ 2}+ b

^{ 2}+ c

^{ 2}) (b

^{ 2}+ c

^{ 2}+ d

^{ 2}) = (ab + bc + cd)

^{ 2}

**Answer**

Let r be the common ratio of the G.P a, b, c, d.

Then b = ar, c = ar

^{2}and d = ar

^{3}

LHS = (a

^{ 2}+ b

^{ 2}+ c

^{ 2}) (b

^{ 2}+ c

^{ 2}+ d

^{ 2})

= (a

^{ 2}+ a

^{ 2}r

^{ 2}+ a

^{ 2}r

^{ 4}) (a

^{ 2}r

^{ 2}+ a

^{ 2}r

^{ 4}+ a

^{ 2}r

^{ 6})

= a

^{ 4 }r

^{ 2}(1 + r

^{ 2 }+ r

^{ 4})

^{ 2}

RHS = (ab + bc +cd)

^{ 2}= (a

^{ 2}r + a

^{ 2}r

^{ 3}+ a

^{ 2}r

^{ 5}) = a

^{ 4}r

^{ 2}(1 + r

^{ 2}+ r

^{ 4})

^{ 2}

Hence, (a

^{ 2}+ b

^{ 2}+ c

^{ 2}) (b

^{ 2}+ c

^{ 2}+ d

^{ 2})

= (ab + bc + cd)

^{2}

26. Insert two number between 3 an 81 so that the resulting sequence is G.P.

**Answer**

Let G

_{1}, G

_{2}be the two numbers such that 3, G

_{1}, G

_{2}, 81 are in G..P.

Let r be the common ratio,

T

_{4}= ar

^{n – 1}= 81 or 3. r

^{ 3}= 81

∴ r

^{3}= 81/3 = 27 = 3

^{3}∴ r = 3

G

_{1}= ar = 3 × 3 = 9; G

_{2}= ar

^{2}= 3 × 3

^{2}= 27

9, 27 are the required number.

27. Find the value of n so that (a

^{n +1 }+

^{bn+1})/(a

^{n}+ b

^{n}) may be the geometric mean between a and b.

**Answer**

The G.M. between a and b = √

*ab*

(a

^{n + 1 }+ b

^{ n+1})/(a

^{n}+ b

^{n}) = √

*ab*= a

^{1/2}b

^{1/2}

Cross – multiplying, we get

a

^{n+1 }+ b

^{n+1 }= a

^{n + 1/2 }b

^{1/2}+ a

^{1/2 }b

^{n+1/2}

cross – multiplying, we get

a

^{n+1 }+ b

^{n+1 }= a

^{n + 1/2}b

^{1/2 }+ a

^{1/2 }b

^{n + 1/2}

=> a

^{n+ 1 }– a

^{n+1}b

^{1/2 }= a

^{1/2 }b

^{n + 1/2 }– b

^{n+ 1}

=> a

^{n + 1/2 }(a

^{1/2 }– b

^{1/2}) = b

^{n + 1/2 }(a

^{1/2 }– b

^{1/2})

Since a and b are different.

Cancelling (a

^{1/2 }– b

^{1/2}) from both sides, we get

a

^{n + 1/2 }= b

^{n + 1/2 }=> (a

^{n + 1/2})/(b

^{n + 1/2}) = 1

=> (a/b)

^{n}

^{+ 1/2 }= 1 = (a/b)

^{0}

n + 1/2 = 0 => n = -(1/2).

28. The sum of two numbers is 6 times their geometric means. Show that numbers are in the ratio (3 + 2√2):(3 - 2√2).

**Answer**

Let numbers be a and b.

Sum of two numbers = 6 × their G.M.

=> a + b = 6√

*ab*=> (a + b)/(2√

*ab*) = 3/1

Applying componendo and dividendo

(a + b + 2√

*ab*)/(a + b – 2√

*ab*) = (3 + 1)/(3 – 1)

=> (√a + √b)

^{2}/(√a – √b)

^{2}= (√4)

^{2}/(√2)

^{2}

=> (√a + √b)/(√a – √b) = 2/√2 = √2/1

Again applying componendo and dividend

((√a + √b) +(√a – √b))/((√a + √b) – (√a – √b)) = (√2 + 1)/(√2 – 1)

=> 2√a/2√b = (√2 + 1)/(√2 – 1) => √a/√b = (√2 + 1)/(√2 – 1)

Squaring we get

a/b = (2 + 1 + 2√2)/(2 + 1 – 2√2) => a/b = (3 + 2√2)/(3 – 2√2).

29. If A and G be A.M and G.M respectively between two positive numbers. Prove that the numbers are A ± √(A + G)(A - G).

**Answer**

30. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2

^{nd}hour, 4

^{th}hour and n

^{ th}hour?

**Answer**

Number of bacteria present in the culture form a G.P.

Whose first term a = 30 and common ratio r = 2

∴ Bacteria present after 2 hours

= ar

^{2}= 30 × 22 = 30 × 4 = 120

Bacteria present after 4 hours = ar

^{4}= 30 × 2

^{4}

= 30 × 16 = 480

Bacteria present after n hours = ar

^{n}= 30 × 2

^{n}

= 30. 2

^{n }

31. What will ₹ 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?

**Answer**

Let A, P, r, T be the amount, principal, rate of interest percent per annum and period in years respectively. The amount A is given by,

A = P(1 + r/100)

^{T}

P = ₹500, r = 10% p.a, T = 10 years;

A = 500(1 + 10/100)

^{10}= 500 × (1.1)

^{10}

32. If A.M. and G.M of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.

**Answer**

Let Î±, Î² be the roots of the quadratic equation.

A.M. of Î±, Î² = (Î± + Î²)/2 = 8;

G.M. ofÎ±,Î² = Î±Î² = 5 => √Î±Î² = 5

^{2}

Equation whose roots are Î±,Î² is x

^{ 2}– (Î± + Î²)

x + Î±Î² = 0

x

^{ 2}– 16x + 25 = 0