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

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^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;
T₂₀ = 5/2(1/2)^{20 – 1} = 5/2 . 1/2¹⁹ = 5/2²⁰
T_n = 5/2 (1/2)^{n -1} = 5/2 . 1/2^n-1 = 5/2ⁿ

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 T₈ = ar^8–1 = ar⁷ = 192 … (i)
Now T₁₂ = ar^{12 – 1} = ar¹¹ = (ar⁷ )r ⁴
= 192 × 2⁴ [ ∵ 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
T₅ = p => ar⁴ = p … (i)
T₈ = q => ar⁷ = q … (ii)
T₁₁ = s => ar¹⁰ = s … (iii)
Now, q² = (ar⁷)² [Using (ii)]
=> q² = a²r¹⁴
=> q² = (ar⁴)(ar¹⁰)
=> q² = 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
T₄ of G.P. = (T₂ of G.P.)² and a = – 3
ar³ = (ar) ²
=> ar³ = a²r²
=> r = a = -3
Now, T₇ = ar⁶ = (–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 = 2⁶
=> (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 = (3²)³ = 3⁶
=> 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 –rⁿ))/(1 – r), r < 1;
S₂₀ = (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(rⁿ – 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, a², – a³, ....... n terms (if a ≠ – 1)

Answer

G.P. is 1, – a, a², – a ³ , .......
Now first term A = 1, r = – a
S_n = (a(1 – rⁿ))/(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 = x³, r= x⁵/x³ = x², x ≠ ± 1
S_n = (a(1 – rⁿ))/(1 – r)
S_n = (x³(1 – x²ⁿ)/(1 – x²)

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
a³ = 1 => a = 1
Sum of these term1/r + 1 + r = 39/10 (put a = 1)
Multiplying by 10r; 10 + 10r + 10r² = 39r
=> 10r² + 10r – 39r + 10 = 0
=> 10r² – 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, 3², 3³, .......... are needed to give the sum 120?

Answer

Let n be the number of terms of the G.P. 3, 3², 3 ³, .......... makes the sum = 120
We have a =3, r = 3
S = (a(rⁿ -1))/(r – 1), r > 1;
Sum = (3(3ⁿ – 1))/(3 – 1) = 120
Or 3/2 (3ⁿ – 1) = 120
Multiplying both sides by 3/2
∴ 3ⁿ – 1 = 80
∴ 3ⁿ = 80 + 1= 81 = 3⁴ => 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.
S₃ = 16
(a(1 – r³))/(1- r) = 16 …….(i)
S₆ – S₃ = 128 => (a(1 – r⁶))/ (1 – r) – 16 = 128,
i.e., (a(1 –r⁶))/(1 – r) = 144 ….(ii)
(ii) ÷ (i) gives (1 – r⁶)/(1 – r³) = 144/16 = 9/1
=> ((1 – r³)(1 + r³))/(1 – r³) = 9/1 => 1 + r³ = 9
=> r³ = 8 => r³ = 2³ => r = 2
Thus, common ratio = 2
As, r = 2, S₃ = (a(r³ – 1))/(r – 1) = 16
=> (a(2³ – 1)/(2 – 1) = 16 => (a(8 – 1))/1 = 16
=> a(7) = 16 => a = 16/7
S_n = (a(rⁿ – 1))/r – 1 = 16/7 . (2ⁿ – 1)/(2 – 1) = 16/7 .(2ⁿ – 1).

15. Given a G.P. with a = 729 and 7^th term 64, determine S₇

Answer

a = 729, T₇ = 64.Let common ratio = r
=> ar⁶ = T₇ = 64 => 729 r ⁶ = 64 => r ⁶ = 64/729
=> r⁶ = (2/3)⁶ => r = 2/3 < 1
S_n = (a(1 – rⁿ))/(1 – r)
S₇ = (729[1 – (2/3)⁷])/1 – 2/3
= 729 × (3(37– 2⁷))/3⁷
= (3⁶.3(3⁷ – 3⁷)/3⁷ = 3⁷ – 2⁷ = 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, T₅ = 4T₃
=> (a(1 – r²))/(1 – r) = -4 => a(1 + r) = -4
=> a( 1 – r) = -4, ar⁴ = 4ar² => r² = 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.
T₄ = x => ar³ = x … (i)
T₁₀ = y => ar⁹ = y … (ii)
T₁₆ = z = ar¹⁵ = z …(iii)
Now, x, y, z will be in G.P.
If ar³, ar⁹, ar¹⁵ are in G.P.
i.e., ar⁹/ar³ = ar¹⁵/ar⁹ => r⁶ = r⁶, 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) + (10² – 1) + (10³ – 1) + …………… to n terms]
= 8/9 [(10 + 10² + 10³ + …………….. to n terms ) – (1 + 1+ 1 ……….. to n terms)]
= 8/9 [(10(10ⁿ – 1))/(10 – 1) - n]
S = 8/9 [10/9 (10ⁿ – 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 –r⁵))/(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, ar² .......... ar^{n – 1} and A, AR, AR², ...... AR^{n – 1} form a G. P, and find the common ratio.

Answer

The two sequences are a, ar, ar² ........ ar^{n – 1} and A, AR, AR², ..........AR^{n – 1}
∴ Sum of the products of the corresponding term of these sequence
aA + aA(rR)+ aA(r²R²) + .......+ 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}, T₂ = ar, T₃ = ar² and T₄ = ar³
Since third term is greater than the first by 9
T₃ = T₁ + 9 => ar² = a + 9 …(i)
Also the second term is greater than the fourth by 18
∴ T₂ = T₄ + 18 => ar = ar³ + 18 …(ii)
Multiplying (i) by r, we get,
ar³ = 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, T₂ =ar = 3(-2) = - 6;
T₃ = ar² = 3(-2)² = 12;
T₄ = ar³ = 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}
=A⁰R⁰ = 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 P² = (ab)ⁿ

Answer

Let a be the first term and r the common ratio of a G.P.
T₁ = a and T_n = b
ar^{n – 1} = b … (i)
Also, P = T₁. T₂. T₃. .............T_n
= a. ar. ar², ..........ar^{n – 1}
= a ⁿ. r ^{1 + 2 + 3 + .........+ (n –1)}
Since, 1 + 2 + 3 + .........+ (n – 1) = (n(n -1))/2
∴ P = aⁿ . r^{n(n – 1)/2}
∴ P² = a²ⁿ . r^{n(n – 1)} = [a² r^n-1]ⁿ
= [a. ar^{n -1}]n = (a . b)ⁿ [using (i)]
Hence, P² = (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/rⁿ

Answer

Sum of the first n terms of a G.P. = a + ar + ar² + ..............+ ar^{n – 1} = (a(1- rn))/(1 – r) ………(i)
 T_{n + 1} = arⁿ, 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
= arⁿ + ar^{n + 1} + ar^{n + 2} + .........+ar^{2n – 1}
= (arⁿ(1 – rⁿ))/(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 = ar² and d = ar³
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 G₁, G₂ be the two numbers such that 3, G₁, G₂, 81 are in G..P.
Let r be the common ratio,
T₄ = ar^{n – 1} = 81 or 3. r ³ = 81
∴ r³ = 81/3 = 27 = 3³ ∴ r = 3
G₁ = ar = 3 × 3 = 9; G₂ = ar² = 3 × 3² = 27
9, 27 are the required number.

27. Find the value of n so that (a^{n +1} + ^bn+1)/(aⁿ + bⁿ) may be the geometric mean between a and b.

Answer

The G.M. between a and b = √ab
(a^{n + 1} + b ⁿ⁺¹)/(aⁿ + bⁿ) = √ab = a^1/2b^1/2
Cross – multiplying, we get
aⁿ⁺¹ + bⁿ⁺¹ = a^{n + 1/2} b^1/2 + a^1/2 b ^n+1/2
cross – multiplying, we get
aⁿ⁺¹ + bⁿ⁺¹ = a ^{n + 1/2} b^1/2 + a^1/2 b ^{n + 1/2}
=> a ^{n+ 1} – aⁿ⁺¹b^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
= ar² = 30 × 22 = 30 × 4 = 120
Bacteria present after 4 hours = ar⁴ = 30 × 2⁴
= 30 × 16 = 480
Bacteria present after n hours = arⁿ = 30 × 2ⁿ
= 30. 2ⁿ

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 α, β be the roots of the quadratic equation.
A.M. of α, β = (α + β)/2 = 8;
G.M. ofα,β = αβ = 5 => √αβ = 5²
Equation whose roots are α,β is x ² – (α + β)
x + αβ = 0
x ² – 16x + 25 = 0