NCERT Solutions for Class 11 Maths Exercise 9.3 Chapter 9 Sequences and Series (2024)

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 8.

Chapter 9 Sequences and Series of Class 11 Maths is categorised under the CBSE Syllabus for 2023-24. NCERT Solutions of this chapter’s exercises are helpful for the students to improve their hold on the problems related to sequences and series. All the questions of this exercise have been solved by subject experts. Exercise 9.3 of NCERT Solutions for Class 11 Maths Chapter 9 – Sequences and Series is based on the following topics:

1. Geometric Progression (G. P.)
   1. The general term of a G.P
   2. Sum to n terms of a G.P
   3. Geometric Mean (G.M.)
2. Relationship between A.M. and G.M.

These solutions are prepared by subject matter experts at BYJU’S, describing the complete method of solving problems. By understanding the concepts used in NCERT Solutions for Class 11 Maths, students will be able to clear all their doubts related to the chapters of Class 11 and ace their board examination.

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

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Access Other Exercise Solutions of Class 11 Maths Chapter 9 – Sequences and Series

Exercise 9.1 Solutions 14 Questions

Exercise 9.2 Solutions 18 Questions

Exercise 9.4 Solutions 10 Questions

Miscellaneous Exercise on Chapter 9 Solutions 32 Questions

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Access Solutions for Class 11 Maths Chapter 9.3 Exercise

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

Solution:

Given G.P. is5/2, 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^thterm 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, letabe the first term of the G.P.

Now,

a₈=ar^8–1=ar⁷

ar⁷= 192

a(2)⁷= 192

a(2)⁷= (2)⁶(3)

3. The 5^th, 8^thand 11^thterms of a G.P. arep,qands, respectively. Show thatq²=ps.

Solution:

Let’s takeato be the first term andrto be the common ratio of the G.P.

Then according to the question, we have

a₅=ar^5–1=ar⁴=p… (i)

a₈=ar^8–1=ar⁷=q… (ii)

a₁₁= ar^11–1=ar¹⁰=s… (iii)

Dividing equation (ii) by (i), we get

4. The 4^th term of a G.P. is the square of its second term, and the first term is –3. Determine its 7^thterm.

Solution:

Let’s considerato be the first term andrto 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= 9r²

⇒r= –3

a₇=ar^7–1=ar⁶= (–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 n^th term of this sequence as 128, we have

Therefore, the 13^th term of the given sequence is 128.

(ii) Given the 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 ofx, 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, forx= ± 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 tonterms 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 tonterms in the geometric progression 1, -a, a², -a³ …. (if a ≠ -1)

Solution:

The given G.P. is1, -a, a², -a³ ….

Here, the first term =a₁= 1

And the common ratio =r= –a

We know that,

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

Solution:

Given G.P. isx³, x⁵, x⁷, …

Here, we havea=x³andr=x⁵/x³ = x²

11. Evaluate:

Solution:

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

Solution:

Leta/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 thatn terms of this G.P. be required to obtain the sum 120.

We know that,

Here,a= 3 andr= 3

Equating the exponents, we get n= 4

Therefore, four terms of the given G.P. are required to obtain the sum 120.

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 tonterms of the G.P.

Solution:

Let’s assume the G.P. to bea,ar,ar²,ar³, …

Then according to the question, we have

a+ar+ar²= 16 andar³+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. witha= 729 and 7^thterm 64, determine S₇.

Solution:

Given,

a= 729 and a₇= 64

Letrbe the common ratio of the G.P.

Then we know that,a_n=a r^n–1

a₇=ar^7–1= (729)r⁶

⇒ 64 = 729r⁶

r⁶ = 64/729

r⁶ = (2/3)⁶

r = 2/3

And we know that

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

Solution:

Considerato be the first term andrto 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^thand 16^thterms of a G.P. arex, yandz, respectively. Prove thatx,y, and zare in G.P.

Solution:

Letabe the first term andrbe the common ratio of the G.P.

According to the given condition,

a₄=ar³=x… (1)

a₁₀=ar⁹=y… (2)

a₁₆=a r¹⁵=z… (3)

On dividing (2) by (1), we get

18. Find the sum tonterms 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 as

S_n= 8 + 88 + 888 + 8888 + …………….. tonterms

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 the 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 sequencesa, ar, ar², …ar^n-1 and A, AR, AR², … AR^n-1form 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 the third term is greater than the first term by 9, and the second term is greater than the 4^thby 18.

Solution:

Considerato be thefirst term andrto 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 ofrin (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^thterms of a G.P. area, bandc, respectively. Prove that a^q-r b^r-p c^p-q = 1

Solution:

Let’s take Ato be the first term andRto 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–rb^r–pc^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 then^thterm of a G.P. area andb, respectively, and ifPis the product ofnterms, prove thatP²= (ab)ⁿ.

Solution:

Given, the first term of the G.P isa,and the last term isb.

Thus,

The G.P. isa,ar,ar²,ar³, …ar^n–1, whereris the common ratio.

Then,

b=ar^n–1 … (1)

P= Product ofnterms

= (a) (ar) (ar²) … (ar^n–1)

= (a×a×…a) (r×r²× …r^n–1)

=aⁿr^{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 the ratio of the sum of the first nterms of a G.P. to the sum of terms from.

Solution:

Letabe the first term andrbe the common ratio of the G.P.

Since there arenterms from (n+1)^thto (2n)^thterm,

Sum of terms from(n+ 1)^thto (2n)^thterm

aⁿ⁺¹=ar^{n + 1– 1}=arⁿ

Thus, the required ratio =

Thus, the ratio of the sum of the first nterms of a G.P. to the sum of terms from (n+ 1)^thto (2n)^thterm is^.

25. Ifa, b, candd are in G.P., show that(a² + b² + c²)(b² + c² + d²) = (ab + bc + cd)².

Solution:

Given, a,b,c,dare 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

=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 assumeG₁andG₂to be two numbers between 3 and 81 such that the series 3,G₁,G₂, 81 forms a G.P.

And letabe the first term andrbe 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)

Forr= 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 ofnso thatmay be the geometric mean betweenaandb.

Solution:

We know that,

The G. M. ofaandbisgiven by √ab.

Then, from the question, we have

By squaring both sides, we get

Performing cross multiplication after expanding, 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 to be aandb.

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 thatAandGare A.M. and G.M. between two positive numbers.

And, let these two positive numbers beaandb.

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^ndhour, 4^thhour andn^thhour?

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 andr= 2

So,a₃=ar²= (30) (2)²= 120

Thus, the number of bacteria at the end of 2^ndhour will be 120.

And, a₅=ar⁴= (30) (2)⁴= 480

The number of bacteria at the end of 4^thhour will be 480.

a_n+1=arⁿ= (30) 2ⁿ

Therefore, the number of bacteria at the end ofn^thhour will be 30(2)ⁿ.

31. What will Rs 500 amount to in 10 years after its deposit in a bank which pays an 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^ndyear, amount = Rs 500 (1.1) (1.1)

At the end of 3^rdyear, 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 beaandb.

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 isx²– 16x+ 25 = 0