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

Chapter 9 Sequences and Series of Class 11 Maths is categorized under the term â€“ I CBSE Syllabus for 2021-22. NCERT Solutions of this chapterâ€™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 BYJUâ€™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 term â€“ I examination.

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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

#### Access Solutions for Class 11 Maths Chapter 9.3 Exercise

1. Find the 20thÂ andÂ nthterms of the G.P. 5/2, 5/4, 5/8, â€¦â€¦â€¦

Solution:

Given G.P. isÂ 5/2, 5/4, 5/8, â€¦â€¦â€¦

Here,Â aÂ = First term =Â 5/2

rÂ = Common ratio =Â (5/4)/(5/2) = Â½

Thus, the 20th term and nth term

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

Solution:

Given,

The common ratio of the G.P.,Â rÂ = 2

And, letÂ aÂ be the first term of the G.P.

Now,

a8Â =Â arÂ 8â€“1Â =Â ar7

ar7Â = 192

a(2)7Â = 192

a(2)7Â = (2)6Â (3)

3. The 5th, 8thÂ and 11thÂ terms of a G.P. areÂ p,Â qÂ andÂ s, respectively. Show thatÂ q2Â =Â ps.

Solution:

Letâ€™s takeÂ aÂ to be the first term andÂ rÂ to be the common ratio of the G.P.

Then according to the question, we have

a5Â =Â aÂ r5â€“1Â =Â aÂ r4Â =Â pÂ â€¦ (i)

a8Â =Â aÂ r8â€“1Â =Â aÂ r7Â =Â qÂ â€¦ (ii)

a11Â = aÂ r11â€“1Â =Â aÂ r10Â =Â sÂ â€¦ (iii)

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

4. The 4thÂ term of a G.P. is square of its second term, and the first term is â€“3. Determine its 7thÂ term.

Solution:

Letâ€™s considerÂ aÂ to be the first term andÂ rÂ to be the common ratio of the G.P.

Given, aÂ = â€“3

And we know that,

anÂ =Â arnâ€“1

So, a4Â =Â ar3Â = (â€“3)Â r3

a2Â =Â a r1Â = (â€“3)Â r

Then from the question, we have

(â€“3)Â r3Â = [(â€“3)Â r]2

â‡’ â€“3r3Â = 9Â r2

â‡’Â rÂ = â€“3

a7Â =Â aÂ rÂ 7â€“1Â =Â aÂ r6Â = (â€“3) (â€“3)6Â = â€“ (3)7Â = â€“2187

Therefore, the seventh term of the G.P. is â€“2187.

5. Which term of the following sequences:

(a) 2, 2âˆš2, 4,â€¦ is 128 ? (b) âˆš3, 3, 3âˆš3,â€¦ is 729 ?

(c) 1/3, 1/9, 1/27, â€¦ is 1/19683 ?

Solution:

(a) The given sequence, 2, 2âˆš2, 4,â€¦

We have,

a = 2 and r = 2âˆš2/2 = âˆš2

Taking the nth term of this sequence as 128, we have

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

(ii) Given sequence, âˆš3, 3, 3âˆš3,â€¦

We have,

a = âˆš3 and r = 3/âˆš3 = âˆš3

Taking the nth term of this sequence to be 729, we have

Therefore, the 12th term of the given sequence is 729.

(iii) Given sequence, 1/3, 1/9, 1/27, â€¦

a = 1/3 and r = (1/9)/(1/3) = 1/3

Taking the nth term of this sequence to be 1/19683, we have

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

6. For what values ofÂ x,Â the numbersÂ -2/7, x, -7/2 are in G.P?

Solution:

The given numbers are -2/7, x, -7/2.

Common ratioÂ = x/(-2/7) = -7x/2

Also, common ratio =Â (-7/2)/x = -7/2x

Therefore, forÂ xÂ = Â± 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 â€¦

Solution:

Given G.P., 0.15, 0.015, 0.00015, â€¦

Here,Â aÂ = 0.15 and r = 0.015/0.15 = 0.1Â

8. Find the sum toÂ nÂ terms in the geometric progression âˆš7, âˆš21, 3âˆš7, â€¦.

Solution:

The given G.P is âˆš7, âˆš21, 3âˆš7, â€¦.

Here,

a = âˆš7 and

9. Find the sum toÂ nÂ terms in the geometric progression 1, -a, a2, -a3 â€¦. (if a â‰  -1)

Solution:

The given G.P. isÂ 1, -a, a2, -a3 â€¦.

Here, the first term =Â a1Â = 1

And the common ratio =Â rÂ = â€“Â a

We know that,

10. Find the sum toÂ nÂ terms in the geometric progression x3, x5, x7, â€¦ (if x â‰  Â±1 )

Solution:

Given G.P. isÂ x3, x5, x7, â€¦

Here, we haveÂ aÂ =Â x3Â andÂ rÂ =Â x5/x3 = x2

11. Evaluate:

Solution:

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

Solution:

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

a/r + a + ar = 39/10 â€¦â€¦ (1)

(a/r) (a) (ar) = 1 â€¦â€¦.. (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 â€“ 29r + 10 = 0

10r2 â€“ 25r â€“ 4r + 10 = 0

5r(2r â€“ 5) â€“ 2(2r â€“ 5) = 0

(5r â€“ 2) (2r â€“ 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, â€¦ are needed to give the sum 120?

Solution:

Given G.P. is 3, 32, 33, â€¦

Letâ€™s consider thatÂ nÂ terms of this G.P. be required to obtain the sum of 120.

We know that,

Here,Â aÂ = 3 andÂ rÂ = 3

Equating the exponents we get, nÂ = 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 toÂ nÂ terms of the G.P.

Solution:

Letâ€™s assume the G.P. to beÂ a,Â ar,Â ar2,Â ar3, â€¦

Then according to the question, we have

aÂ +Â arÂ +Â ar2Â = 16 andÂ ar3Â +Â ar4Â +Â ar5Â = 128

aÂ (1 +Â rÂ +Â r2) = 16 â€¦ (1) and,

ar3(1 +Â rÂ +Â r2) = 128 â€¦ (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. withÂ aÂ = 729 and 7thÂ term 64, determine S7.

Solution:

Given,

aÂ = 729 and a7Â = 64

LetÂ rÂ be the common ratio of the G.P.

Then we know that,Â anÂ =Â a rnâ€“1

a7Â =Â ar7â€“1Â = (729)r6

â‡’ 64 = 729Â 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 â€“4 and the fifth term is 4 times the third term.

Solution:

ConsiderÂ aÂ to be the first term andÂ rÂ 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 = Â± 2

Using the value of r in (1), we have

Therefore, the required G.P is

-4/3, -8/3, -16/3, â€¦. Or 4, -8, 16, -32, â€¦â€¦

17. If the 4th, 10thÂ and 16thÂ terms of a G.P. areÂ x, yÂ andÂ z, respectively. Prove thatÂ x,Â y,Â zÂ are in G.P.

Solution:

LetÂ aÂ be the first term andÂ rÂ be the common ratio of the G.P.

According to the given condition,

a4Â =Â aÂ r3Â =Â xÂ â€¦ (1)

a10Â =Â aÂ r9Â =Â yÂ â€¦ (2)

a16Â =Â a r15Â =Â zÂ â€¦ (3)

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

18. Find the sum toÂ nÂ terms of the sequence, 8, 88, 888, 8888â€¦

Solution:

Given sequence: 8, 88, 888, 8888â€¦

This sequence is not a G.P.

But, it can be changed to G.P. by writing the terms as

SnÂ = 8 + 88 + 888 + 8888 + â€¦â€¦â€¦â€¦â€¦.. toÂ nÂ 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,Â 1/2.

Solution:

The required sum =Â 2 x 128 + 4 x 32 + 8 x 8 + 16 x 2 + 32 x Â½

= 64[4 + 2 + 1 + Â½ + 1/22]

Now, itâ€™s seen that

4, 2, 1,Â Â½, 1/22 is a G.P.

With first term,Â aÂ = 4

Common ratio,Â rÂ =1/2

We know,

Therefore, the required sum =Â 64(31/4) = (16)(31) = 496

20. Show that the products of the corresponding terms of the sequencesÂ a, ar, ar2, â€¦arn-1 and A, AR, AR2, â€¦ ARn-1Â form a G.P, and find the common ratio.

Solution:

To be proved: The sequence,Â aA,Â arAR,Â ar2AR2, â€¦arnâ€“1ARnâ€“1, forms a G.P.

Now, we have

Therefore, the above sequence forms a G.P. and the common ratio isÂ 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 4thÂ by 18.

Solution:

ConsiderÂ aÂ to be theÂ first term andÂ rÂ to be the common ratio of the G.P.

Then,

a1Â =Â a,Â a2Â =Â ar,Â a3Â =Â ar2,Â a4Â =Â ar3

From the question, we have

a3Â =Â a1Â + 9

ar2Â =Â aÂ + 9 â€¦ (i)

a2Â =Â a4Â + 18

arÂ =Â ar3Â + 18 â€¦ (ii)

So, from (1) and (2), we get

a(r2Â â€“ 1) = 9 â€¦ (iii)

arÂ (1â€“Â r2) = 18 â€¦ (iv)

Now, dividing (4) by (3), we get

-r = 2

r = -2

On substituting the value ofÂ rÂ in (i), we get

4aÂ =Â aÂ + 9

3aÂ = 9

âˆ´Â aÂ = 3

Therefore, the first four numbers of the G.P. are 3, 3(â€“ 2), 3(â€“2)2, and 3(â€“2)3

i.e., 3Â¸â€“6, 12, and â€“24.

22. If the pth, qth and rthÂ terms of a G.P. areÂ a, bÂ andÂ c, respectively. Prove that aq-r br-p cp-q = 1

Solution:

Letâ€™s take AÂ to be the first term andÂ RÂ to be the common ratio of the G.P.

Then according to the question, we have

ARpâ€“1Â =Â a

ARqâ€“1Â =Â b

ARrâ€“1Â =Â c

Then,

aqâ€“rÂ brâ€“pÂ cpâ€“q

=Â Aqâ€“rÂ Ã—Â R(pâ€“1) (qâ€“r)Â Ã— Arâ€“pÂ Ã—Â R(qâ€“1) (râ€“p)Â Ã—Â Apâ€“qÂ Ã—Â R(rÂ â€“1)(pâ€“q)

=Â AqÂ â€“Â rÂ +Â rÂ â€“Â pÂ +Â pÂ â€“Â qÂ Ã—Â RÂ (prÂ â€“Â prÂ â€“Â qÂ +Â r) + (rqÂ â€“Â rÂ +Â pÂ â€“Â pq) + (prÂ â€“Â pÂ â€“Â qrÂ +Â q)

=Â A0Â Ã—Â R0

= 1

Hence proved.

23. If the first and theÂ nthÂ term of a G.P. areÂ aÂ adÂ b, respectively, and ifÂ PÂ is the product ofÂ nÂ terms, prove thatÂ P2Â = (ab)n.

Solution:

Given, the first term of the G.P isÂ aÂ and the last term isÂ b.

Thus,

The G.P. isÂ a,Â ar,Â ar2,Â ar3, â€¦Â arnâ€“1, whereÂ rÂ is the common ratio.

Then,

bÂ =Â arnâ€“1Â  â€¦ (1)

PÂ = Product ofÂ nÂ terms

= (a) (ar) (ar2) â€¦ (arnâ€“1)

= (aÂ Ã—Â aÂ Ã—â€¦a) (rÂ Ã—Â r2Â Ã— â€¦rnâ€“1)

=Â anÂ rÂ 1 + 2 +â€¦(nâ€“1)Â  â€¦ (2)

Here, 1, 2, â€¦(nÂ â€“ 1) is an A.P.

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

24. Show that the ratio of the sum of firstÂ nÂ terms of a G.P. to the sum of terms fromÂ .

Solution:

LetÂ aÂ be the first term andÂ rÂ be the common ratio of the G.P.

Since there areÂ nÂ terms from (nÂ +1)thÂ to (2n)thÂ term,

Sum of terms from(nÂ + 1)thÂ to (2n)thÂ term

aÂ nÂ +1Â =Â arÂ n + 1Â â€“ 1Â =Â arn

Thus, required ratio =

Thus, the ratio of the sum of firstÂ nÂ terms of a G.P. to the sum of terms from (nÂ + 1)thÂ to (2n)thÂ term isÂ .

25. IfÂ a, b, cÂ andÂ dÂ are in G.P. show thatÂ (a2 + b2 + c2)(b2 + c2 + d2) = (ab + bc + cd)2.

Solution:

Given, a,Â b,Â c,Â dÂ are in G.P.

So, we have

b2Â =Â acÂ  â€¦ (2)

c2Â =Â bdÂ  â€¦ (3)

Taking the R.H.S. we have

R.H.S.

= (abÂ +Â bcÂ +Â cd)2

= (abÂ +Â adÂ +Â cd)2Â  [Using (1)]

= [abÂ +Â dÂ (aÂ +Â c)]2

=Â a2b2Â + 2abdÂ (aÂ +Â c) +Â d2Â (aÂ +Â c)2

=Â a2b2Â +2a2bdÂ + 2acbdÂ +Â d2(a2Â + 2acÂ +Â c2)

=Â a2b2Â + 2a2c2Â + 2b2c2Â +Â d2a2Â + 2d2b2Â +Â d2c2Â  [Using (1) and (2)]

=Â a2b2Â +Â a2c2Â +Â a2c2Â +Â b2c2Â +Â b2c2Â +Â d2a2Â +Â d2b2Â +Â d2b2Â +Â d2c2

=Â a2b2Â +Â a2c2Â +Â a2d2Â +Â b2Â Ã—Â b2Â +Â b2c2Â +Â b2d2Â +Â c2b2Â +Â c2Â Ã—Â c2Â +Â c2d2

[Using (2) and (3) and rearranging terms]

=Â a2(b2Â +Â c2Â +Â d2) +Â b2Â (b2Â +Â c2Â +Â d2) +Â c2Â (b2+Â c2Â +Â d2)

= (a2Â +Â b2Â +Â c2) (b2Â +Â c2Â +Â 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:

Letâ€™s assumeÂ G1Â andÂ G2Â to be two numbers between 3 and 81 such that the series 3,Â G1,Â G2, 81 forms a G.P.

And letÂ aÂ be the first term andÂ rÂ be the common ratio of the G.P.

Now, we have the 1st term as 3 and the 4th term as 81.

81 = (3)Â (r)3

r3Â = 27

âˆ´Â rÂ = 3 (Taking real roots only)

ForÂ rÂ = 3,

G1Â =Â arÂ = (3) (3) = 9

G2Â =Â ar2Â = (3) (3)2Â = 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 ofÂ nÂ so thatÂ may be the geometric mean betweenÂ aÂ andÂ b.

Solution:

We know that,

The G. M. ofÂ aÂ andÂ bÂ isÂ given by âˆšab.

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 beÂ aÂ andÂ b.

Then, G.M. =Â âˆšab.

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 thatÂ AÂ andÂ GÂ are A.M. and G.M. between two positive numbers.

And, let these two positive numbers beÂ aÂ andÂ 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 2ndÂ hour, 4thÂ hour andÂ nthÂ 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,Â aÂ = 30 andÂ rÂ = 2

So,Â a3Â =Â ar2Â = (30) (2)2Â = 120

Thus, the number of bacteria at the end of 2ndÂ hour will be 120.

And, a5Â =Â ar4Â = (30) (2)4Â = 480

The number of bacteria at the end of 4thÂ hour will be 480.

anÂ +1Â =Â arnÂ = (30) 2n

Therefore, the number of bacteria at the end ofÂ nthÂ 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)Â = Rs 500 (1.1)

At the end of 2ndÂ year, amount = Rs 500 (1.1) (1.1)

At the end of 3rdÂ year, amount = Rs 500 (1.1) (1.1) (1.1) and so onâ€¦.

Therefore,

The amount at the end of 10 years = Rs 500 (1.1) (1.1) â€¦ (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:

Letâ€™s consider the roots of the quadratic equation to beÂ aÂ andÂ b.

Then, we have

We know that,

A quadratic equation can be formed as,

x2 â€“Â xÂ (Sum of roots) + (Product of roots) = 0

x2Â â€“Â xÂ (aÂ +Â b) + (ab) = 0

x2Â â€“ 16xÂ + 25 = 0 [Using (1) and (2)]

Therefore, the required quadratic equation isÂ x2Â â€“ 16xÂ + 25 = 0

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