Ncert Solutions Class 11 Maths Chapter 9 Exercise 9.3

Chapter 9 Sequences and Series of Class 11 Maths is categorized under the term – I CBSE Syllabus for 2021-22. 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 term – I examination.

Download PDF of NCERT Solutions for Class 11 Maths Chapter 9- Sequences and Series Exercise 9.3

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 20^th andn ^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, leta be the first term of the G.P.

Now,

a ₈ =ar^8–1 =ar ⁷

ar ⁷ = 192

a(2)⁷ = 192

a(2)⁷ = (2)⁶ (3)

3. The 5^th, 8^th and 11^th terms of a G.P. arep,q ands, respectively. Show thatq ² =ps.

Solution:

Let's takea to be the first term and r to be the common ratio of the G.P.

Then according to the question, we have

a ₅ =ar ^5–1=a r⁴ =p … (i)

a ₈=ar ^8–1=ar ⁷ =q … (ii)

a ₁₁ = ar ^11–1=ar ¹⁰ =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 considera to be the first term andr to be the common ratio of the G.P.

Given, a = –3

And we know that,

a_n  =ar^{n –1}

So, a ₄=ar ³ = (–3)r ³

a ₂ =a r ¹ = (–3)r

Then from the question, we have

(–3)r ³ = [(–3)r]²

⇒ –3r ³ = 9r ²

⇒r = –3

a ₇ =ar^7–1=ar ⁶ = (–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 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, 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 ton 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 ton 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 =a ₁ = 1

And the common ratio =r = –a

We know that,

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

Solution:

Given G.P. is x³, x⁵, x⁷, …

Here, we havea =x ³ andr =x ⁵/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 thatn terms of this G.P. be required to obtain the sum of 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 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 bea,ar,ar ²,ar ³, …

Then according to the question, we have

a+ar +ar ² = 16 andar ³+ar ⁴ +ar ⁵= 128

a (1 +r +r ²) = 16 … (1) and,

ar ³(1 +r +r ²) = 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^th term 64, determine S₇.

Solution:

Given,

a = 729 and a ₇ = 64

Letr be the common ratio of the G.P.

Then we know that,a_n  =a r^{n –1}

a ₇ =ar ^7–1 = (729)r ⁶

⇒ 64 = 729r ⁶

r ⁶ = 64/729

r ⁶ = (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:

Considera to be the first term andr 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. arex, yandz, respectively. Prove thatx, y, zare in G.P.

Solution:

Leta be the first term andr be the common ratio of the G.P.

According to the given condition,

a ₄ =ar ³ =x … (1)

a ₁₀ =ar ⁹ = y … (2)

a ₁₆ = a r ¹⁵=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 + …………….. ton 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,ar ² AR ², …ar^{n –1} AR^{n –1}, forms a G.P.

Now, we have

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

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:

Considera to be thefirst term andr to be the common ratio of the G.P.

Then,

a ₁ =a,a ₂ =ar,a ₃ =ar ²,a ₄ =ar ³

From the question, we have

a ₃ =a ₁ + 9

ar ² =a + 9 … (i)

a ₂ =a ₄ + 18

ar=ar ³ + 18 … (ii)

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

a(r ² – 1) = 9 … (iii)

ar(1–r ²) = 18 … (iv)

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

-r = 2

r = -2

On substituting the value ofr 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. area, b andc, respectively. Prove that a^q-r b^r-p c^p-q = 1

Solution:

Let's take A to be the first term andR 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 )}

=A ⁰ ×R ⁰

= 1

Hence proved.

23. If the first and then ^th term of a G.P. area adb, respectively, and ifP is the product ofn terms, prove thatP ² = (ab) ⁿ .

Solution:

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

Thus,

The G.P. isa,ar,ar ²,ar ³, …ar^{n –1}, wherer is the common ratio.

Then,

b =ar^{n –1}  … (1)

P = Product ofn terms

= (a) (ar) (ar ²) … (ar^{n –1})

= (a ×a ×…a) (r ×r ² × …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 firstn terms of a G.P. to the sum of terms from.

Solution:

Leta be the first term andrbe the common ratio of the G.P.

Since there aren 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 firstn terms of a G.P. to the sum of terms from (n + 1)^th to (2n)^thterm is^.

25. Ifa, b, c andd 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)

b ² =ac … (2)

c ² =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

=a ² b ² + 2abd (a +c) +d ² (a +c)²

=a ² b ² +2a ² bd + 2acbd +d ²(a ² + 2ac +c ²)

=a ² b ² + 2a ² c ² + 2b ² c ² +d ² a ² + 2d ² b ² +d ² c ²  [Using (1) and (2)]

=a ² b ² +a ² c ² +a ² c ² +b ² c ²+b ² c ² +d ² a ² +d ² b ² +d ² b ² +d ² c ²

=a ² b ² +a ² c ² +a ² d ²+b ²×b ² +b ² c ² +b ² d ² +c ² b ² +c ²×c ² +c ² d ²

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

=a ²(b ² +c ² +d ²) +b ² (b ² +c ² +d ²) +c ² (b ²+c ² +d ²)

= (a ² +b ² +c ²) (b ² +c ² +d ²)

= 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 assumeG ₁ andG ₂ to be two numbers between 3 and 81 such that the series 3,G ₁,G ₂, 81 forms a G.P.

And leta be the first term andr 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

r ³ = 27

∴r = 3 (Taking real roots only)

Forr = 3,

G ₁ =ar = (3) (3) = 9

G ₂ =ar ² = (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 thatmay be the geometric mean betweena andb.

Solution:

We know that,

The G. M. ofa andb is given 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 bea andb.

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 andG are A.M. and G.M. between two positive numbers.

And, let these two positive numbers bea andb.

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 andn ^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 andr = 2

So,a ₃ =ar ² = (30) (2)² = 120

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

And, a ₅ =ar ⁴ = (30) (2)⁴ = 480

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

a_{n  +1}=arⁿ  = (30) 2 ⁿ

Therefore, the number of bacteria at the end ofn ^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 bea andb.

Then, we have

We know that,

A quadratic equation can be formed as,

x ² –x (Sum of roots) + (Product of roots) = 0

x ² –x (a +b) + (ab) = 0

x ² – 16x + 25 = 0 [Using (1) and (2)]

Therefore, the required quadratic equation isx ² – 16x + 25 = 0

Ncert Solutions Class 11 Maths Chapter 9 Exercise 9.3

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