NCERT Solutions Class 11th Maths Chapter – 9 Sequences and Series Exercise 9.3

NCERT Solutions Class 11th Maths Chapter – 9 Sequences and Series Exercise 9.3

NCERT Solutions Class 11th Maths Chapter – 9 Sequences and Series Exercise 9.3

?Chapter – 9?

✍Sequences and Series✍

?Exercise 9.3?

1. Find the 20^th and n^thterms of the G.P. 5/2, 5/4, 5/8, ………

?‍♂️solution – Given G.P. is 5/2, 5/4, 5/8, ………
Here, a = First term = 5/2
r = Common ratio = (5/4)/(5/2) = ½
Thus, the 20^th term and n^th term

2. Find the 12^th term of a G.P. whose 8^th term is 192 and the common ratio is 2.

?‍♂️solution – Given,
The common ratio of the G.P., r = 2
And, let a be the first term of the G.P.

3. The 5^th, 8^th and 11^th terms of a G.P. are p, q and s, respectively. Show that q² = ps.

?‍♂️solution – Let’s take a to be the first term and r to be the common ratio of the G.P.
Then according to the question, we have
a₅ = a r^5–1 = a r⁴ = p … (i)
a₈ = a r^8–1 = a r⁷ = q … (ii)
a₁₁ = a r^11–1 = a r¹⁰ = s … (iii)
Dividing equation (ii) by (i), we get

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.

?‍♂️solution – Let’s consider a to be the first term and r to be the common ratio of the G.P.
Given, a = –3
And we know that,
a_n = ar^n–1
So, a₄ = ar³ = (–3) r³
a₂ = a r¹ = (–3) r
Then from the question, we have
(–3) r³ = [(–3) r]²
⇒ –3r³ = 9 r²
⇒ r = –3
a₇ = a r ^7–1 = a r⁶ = (–3) (–3)⁶ = – (3)⁷ = –2187
Therefore, the seventh term of the G.P. is –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 ?

?‍♂️solution – (a) The given sequence, 2, 2√2, 4,…
We have,
a = 2 and r = 2√2/2 = √2
Taking the nth term of this sequence as 128, we have

Therefore, the  13^th term of the given sequence is 128.
(ii) Given sequence, √3, 3, 3√3,…
We have,
a = √3 and r = 3/√3 = √3
Taking the  n^th term of this sequence to be 729, we have

Therefore, the 12^th term of the given sequence is 729.
(iii) Given sequence, 1/3, 1/9, 1/27, …
a = 1/3 and r = (1/9)/(1/3) = 1/3
Taking the n^th term of this sequence to be 1/19683, we have

Therefore, the 9^th term of the given sequence is 1/19683.

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

?‍♂️solution – The given numbers are -2/7, x, -7/2.
Common ratio = x/(-2/7) = -7x/2
Also, common ratio = (-7/2)/x = -7/2x

Therefore, for x = ± 1, the given numbers will be in G.P.

7. Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015 …

?‍♂️solution – Given G.P., 0.15, 0.015, 0.00015, …
Here, a = 0.15 and r = 0.015/0.15 = 0.1 

8. Find the sum to n terms in the geometric progression √7, √21, 3√7, ….

?‍♂️solution – The given G.P is √7, √21, 3√7, ….
Here,
a = √7 and

9. Find the sum to n terms in the geometric progression 1, -a, a², -a³ …. (if a ≠ -1)

?‍♂️solution – The given G.P. is 1, -a, a², -a³ ….
Here, the first term = a1 = 1
And the common ratio = r = – a
We know that,

10. Find the sum to n terms in the geometric progression x³, x⁵, x⁷, … (if x ≠ ±1 )

?‍♂️solution – Given G.P. is x³, x⁵, x⁷, …
Here, we have a = x³ and r = x⁵/x³ = x²

11. Evaluate –

?‍♂️solution – 

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.

?‍♂️solution – Let a/r, a, ar be the first three terms of the G.P.
a/r + a + ar = 39/10 …… (1)
(a/r) (a) (ar) = 1 …….. (2)
From (2), we have
a³ = 1
Hence, a = 1 [Considering real roots only]
Substituting the value of a in (1), we get
1/r + 1 + r = 39/10
(1 + r + r²)/r = 39/10
10 + 10r + 10r² = 39r
10r2 – 29r + 10 = 0
10r2 – 25r – 4r + 10 = 0
5r(2r – 5) – 2(2r – 5) = 0
(5r – 2) (2r – 5) = 0
Thus,
r = 2/5 or 5/2
Therefore, the three terms of the G.P. are 5/2, 1 and 2/5.

13. How many terms of G.P. 3, 3², 3³, … are needed to give the sum 120?

?‍♂️solution – Given G.P. is 3, 3², 3³, …
Let’s consider that n terms of this G.P. be required to obtain the sum of 120.
We know that,

Here, a = 3 and r = 3

Equating the exponents we get, n = 4
Therefore, four terms of the given G.P. are required to obtain the sum as 120.

14. The sum of 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.

?‍♂️solution – Let’s assume the G.P. to be a, ar, ar², ar³, …
Then according to the question, we have
a + ar + ar² = 16 and ar³ + ar⁴ + ar⁵ = 128
a (1 + r + r²) = 16 … (1) and,
ar³(1 + r + r²) = 128 … (2)
Dividing equation (2) by (1), we get

r³ = 8
r = 2
Now, using r = 2 in (1), we get
a (1 + 2 + 4) = 16
a (7) = 16
a = 16/7
Now, the sum of terms is given as

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

?‍♂️solution – Given,
a = 729 and a₇ = 64
Let r be the common ratio of the G.P.
Then we know that, a_n = a r^n–1
a₇ = ar^7–1 = (729)r⁶
⇒ 64 = 729 r⁶
r⁶ = 64/729
r⁶ = (2/3)⁶
r = 2/3
And, we know that

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.

?‍♂️solution – Consider a to be the first term and r to be the common ratio of the G.P.
Given, S₂ = -4
Then, from the question we have

And,
a₅ = 4 x a₃
ar⁴ = 4ar²
r² = 4
r = ± 2
Using the value of r in (1), we have

Therefore, the required G.P is
-4/3, -8/3, -16/3, …. Or 4, -8, 16, -32, ……

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.

?‍♂️solution – Let a be the first term and r be the common ratio of the G.P.
According to the given condition,
a₄ = a r³ = x … (1)
a₁₀ = a r⁹ = y … (2)
a₁₆ = a r¹⁵ = z … (3)
On dividing (2) by (1), we get

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

?‍♂️solution – Given sequence: 8, 88, 888, 8888…
This sequence is not a G.P.
But, it can be changed to G.P. by writing the terms as
S_n  = 8 + 88 + 888 + 8888 + …………….. to n terms

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.

?‍♂️solution – The required sum = 2 x 128 + 4 x 32 + 8 x 8 + 16 x 2 + 32 x ½
= 64[4 + 2 + 1 + ½ + 1/2²]
Now, it’s seen that
4, 2, 1, ½, 1/2² is a G.P.
With first term, a = 4
Common ratio, r =1/2
We know,

Therefore, the required sum = 64(31/4) = (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.

?‍♂️solution – To be proved: The sequence, aA, arAR, ar²AR², …ar^n–1AR^n–1, forms a G.P.
Now, we have

Therefore, the above sequence forms a G.P. and the common ratio is rR.

21. Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4^th by 18.

?‍♂️solution – Consider a to be the first term and r to be the common ratio of the G.P.
Then,
a₁ = a, a₂ = ar, a₃ = ar², a₄ = ar³
From the question, we have
a₃ = a₁ + 9
ar² = a + 9 … (i)
a₂ = a₄ + 18
ar = ar³ + 18 … (ii)
So, from (1) and (2), we get
a(r² – 1) = 9 … (iii)
ar (1– r²) = 18 … (iv)
Now, dividing (4) by (3), we get

-r = 2
r = -2
On substituting the value of r in (i), we get
4a = a + 9
3a = 9
∴ a = 3
Therefore, the first four numbers of the G.P. are 3,3(– 2), 3(–2)², and 3(–2)³
i.e., 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

?‍♂️solution – Let’s take A to be the first term and R to be the common ratio of the G.P.
Then according to the question, we have
AR^p–1 = a
AR^q–1 = b
AR^r–1 = c
Then,
a^q–r b^r–p c^p–q
= A^q–r × R^{(p–1) (q–r)} × A^r–p × R^{(q–1) (r–p)} × A^p–q × R^{(r –1)(p–q)}
= Aq ^{– r + r – p + p – q} × R ^{(pr – pr – q + r) + (rq – r + p – pq) + (pr – p – qr + q)}
=  A⁰ × R⁰
= 1
Hence proved.

23. If the first and the n^th term of a G.P. are a ad b, respectively, and if P is the product of n terms, prove that P² = (ab)ⁿ.

?‍♂️solution – Given, the first term of the G.P is a and the last term is b.
Thus,
The G.P. is a, ar, ar², ar³, … ar^n–1, where r is the common ratio.
Then,
b = ar^n–1  … (1)
P = Product of n terms
= (a) (ar) (ar²) … (ar^n–1)
= (a × a ×…a) (r × r² × …r^n–1)
= aⁿ r ^{1 + 2 +…(n–1)}  … (2)
Here, 1, 2, …(n – 1) is an A.P.

And, the product of n terms P is given by,

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

?‍♂️solution – Let a be the first term and r be the common ratio of the G.P.

Since there are n terms from (n +1)^th to (2n)^th term,
Sum of terms from(n + 1)^th to (2n)^th term

a ^n +1 = ar ^{n + 1 – 1} = arⁿ

Thus, required ratio =

Thus, 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

25. If a, b, c and d are in G.P. show that (a² + b² + c²)(b² + c² + d²) = (ab + bc + cd)².

?‍♂️solution – Given, a, b, c, d are in G.P.
So, we have
bc = ad  … (1)
b² = ac  … (2)
c² = bd  … (3)
Taking the R.H.S. we have
R.H.S.
= (ab + bc + cd)²
= (ab + ad + cd)²  [Using (1)]
= [ab + d (a + c)]²
= a²b² + 2abd (a + c) + d² (a + c)²
= a²b² +2a²bd + 2acbd + d²(a² + 2ac + c²)
= a²b² + 2a²c² + 2b²c² + d²a² + 2d²b² + d²c²  [Using (1) and (2)]
= a²b² + a²c² + a²c² + b²c² + b²c² + d²a² + d²b² + d²b² + d²c²
= a²b² + a²c² + a²d² + b² × b² + b²c² + b²d² + c²b² + c² × c² + c²d²
[Using (2) and (3) and rearranging terms]
= a²(b² + c² + d²) + b² (b² + c² + d²) + c² (b²+ c² + d²)
= (a² + b² + c²) (b² + c² + d²)
= L.H.S.
Thus, L.H.S. = R.H.S.
Therefore,
(a² + b² + c²)(b² + c² + d²) = (ab + bc + cd)²

26. Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

?‍♂️solution – Let’s assume G₁ and G₂ to be two numbers between 3 and 81 such that the series 3, G₁, G₂, 81 forms a G.P.
And let a be the first term and r be the common ratio of the G.P.
Now, we have the 1^st term as 3 and the 4^th term as 81.
81 = (3) (r)³
r³ = 27
∴ r = 3 (Taking real roots only)
For r = 3,
G₁ = ar = (3) (3) = 9
G₂ = ar² = (3) (3)² = 27
Therefore, the two numbers which can be inserted between 3 and 81 so that the resulting sequence becomes a G.P are 9 and 27.

27. Find the value of n so that may be the geometric mean between a and b.

?‍♂️solution – We know that,
The G. M. of a and b is given by √ab.
Then from the question, we have

By squaring both sides, we get

28. The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio

?‍♂️solution – Consider the two numbers be a and b.
Then, G.M. = √ab.
From the question, we have

29. If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are .

?‍♂️solution – Given that A and G are A.M. and G.M. between two positive numbers.
And, let these two positive numbers be a and b.

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?

?‍♂️solution – Given, the number of bacteria doubles every hour. Hence, the number of bacteria after every hour will form a G.P.
Here we have, a = 30 and r = 2
So, a₃ = ar² = (30) (2)² = 120
Thus, the number of bacteria at the end of 2^nd hour will be 120.
And, a₅ = ar⁴ = (30) (2)⁴ = 480
The number of bacteria at the end of 4^th hour will be 480.
a_n +1 = arⁿ = (30) 2ⁿ
Therefore, the number of bacteria at the end of n^th hour will be 30(2)ⁿ.

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

?‍♂️solution – Given,
The amount deposited in the bank is Rs 500.
At the end of first year, amount = Rs 500(1 + 1/10) = Rs 500 (1.1)
At the end of 2^nd year, amount = Rs 500 (1.1) (1.1)
At the end of 3^rd year, amount = Rs 500 (1.1) (1.1) (1.1) and so on….
Therefore,
The amount at the end of 10 years = Rs 500 (1.1) (1.1) … (10 times)
= Rs 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.

?‍♂️solution – Let’s consider the roots of the quadratic equation to be a and b.
Then, we have

We know that,
A quadratic equation can be formed as,
x² – x (Sum of roots) + (Product of roots) = 0
x²  – x (a + b) + (ab) = 0
x²  – 16x + 25 = 0 [Using (1) and (2)]
Therefore, the required quadratic equation is x²  – 16x + 25 = 0