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

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

Page No 192:

Question 1:

Find the 20^th and n^thterms of the G.P.

Answer:

The given G.P. is 

Here, a = First term = 

r = Common ratio = 

Question 2:

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

Answer:

Common ratio, r = 2

Let a be the first term of the G.P.

∴ a₈ = ar ^8–1 = ar⁷

⇒ ar⁷ = 192

a(2)⁷ = 192

a(2)⁷ = (2)⁶ (3)

Question 3:

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

Answer:

Let a be the first term and r be the common ratio of the G.P.

According to the given condition,

a₅ = a r^5–1 = a r⁴ = p … (1)

a₈ = a r^8–1 = a r⁷ = q … (2)

a₁₁ = a r^11–1 = a r¹⁰ = s … (3)

Dividing equation (2) by (1), we obtain

Dividing equation (3) by (2), we obtain

Equating the values of r³ obtained in (4) and (5), we obtain

Thus, the given result is proved.

Question 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 be the common ratio of the G.P.

∴ a = –3

It is known that, a_n = ar^n–1

∴a₄ = ar³ = (–3) r³

a₂ = a r¹ = (–3) r

According to the given condition,

(–3) r³ = [­(–3) r]²

⇒ –3r³ = 9 r²

⇒ r = –3

a₇ = a r ^7–1 = a r⁶ = (–3) (–3)⁶ = – (3)⁷ = –2187

Thus, the seventh term of the G.P. is –2187.

Question 5:

Which term of the following sequences:

(a)  (b)  (c) 

Answer:

(a) The given sequence is 

Here, a = 2 and r = 

Let the n^th term of the given sequence be 128.

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

(b) The given sequence is 

Here, 

Let the n^th term of the given sequence be 729.

Thus, the 12^th term of the given sequence is 729.

(c) The given sequence is 

Here, 

Let the n^th term of the given sequence be .

Thus, the 9^th term of the given sequence is .

Question 6:

For what values of x, the numbers are in G.P?

Answer:

The given numbers are .

Common ratio = 

Also, common ratio = 

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

Question 7:

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

Answer:

The given G.P. is 0.15, 0.015, 0.00015, …

Here, a = 0.15 and 

Question 8:

Find the sum to n terms in the geometric progression

Answer:

The given G.P. is 

Here, 

Question 9:

Find the sum to n terms in the geometric progression

Answer:

The given G.P. is 

Here, first term = a₁ = 1

Common ratio = r = – a

Question 10:

Find the sum to n terms in the geometric progression

Answer:

The given G.P. is 

Here, a = x³ and r = x²

Question 11:

Evaluate

Answer:

The terms of this sequence 3, 3², 3³, … forms a G.P.

Substituting this value in equation (1), we obtain

Question 12:

The sum of first three terms of a G.P. is and their product is 1. Find the common ratio and the terms.

Answer:

Let be the first three terms of the G.P.

From (2), we obtain

a³ = 1

⇒ a = 1 (Considering real roots only)

Substituting a = 1 in equation (1), we obtain

Thus, the three terms of G.P. are .

Question 13:

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

Answer:

The given G.P. is 3, 3², 3³, …

Let n terms of this G.P. be required to obtain the sum as 120.

Here, a = 3 and r = 3

∴ n = 4

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

Question 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.

Answer:

Let the G.P. be a, ar, ar², ar³, …

According to the given condition,

a + ar + ar² = 16 and ar³ + ar⁴ + ar⁵ = 128

⇒ a (1 + r + r²) = 16 … (1)

ar³(1 + r + r²) = 128 … (2)

Dividing equation (2) by (1), we obtain

Substituting r = 2 in (1), we obtain

a (1 + 2 + 4) = 16

⇒ a (7) = 16

Question 15:

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

Answer:

a = 729

a₇ = 64

Let r be the common ratio of the G.P.

It is known that, a_n = a r^n–1

a₇ = ar^7–1 = (729)r⁶

⇒ 64 = 729 r⁶

Also, it is known that, 

Question 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 of the G.P.

According to the given conditions,

a₅ = 4 × a₃

ar⁴ = 4ar²

⇒ r² = 4

∴ r = ± 2

From (1), we obtain

Thus, the required G.P. is

 4, –8, 16, –32, …

Question 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 of the G.P.

According to the given condition,

a₄ = a r³ = x … (1)

a₁₀ = a r⁹ = y … (2)

a₁₆ = a r¹⁵ = z … (3)

Dividing (2) by (1), we obtain

Dividing (3) by (2), we obtain

∴ 

Thus, x, y, z are in G. P.

Page No 193:

Question 18:

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

Answer:

The given sequence is 8, 88, 888, 8888…

This sequence is not a G.P. However, it can be changed to G.P. by writing the terms as

S_n = 8 + 88 + 888 + 8888 + …………….. to n terms

Question 19:

Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, .

Answer:

Required sum = 

Here, 4, 2, 1, is a G.P.

First term, a = 4

Common ratio, r =

It is known that, 

∴Required sum = 

Question 20:

Show that the products of the corresponding terms of the sequences  form a G.P, and find the common ratio.

Answer:

It has to be proved that the sequence, aA, arAR, ar²AR², …ar^n–1AR^n–1, forms a G.P.

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

Question 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.

Answer:

Let a be the first term and r be the common ratio of the G.P.

a₁ = a, a₂ = ar, a₃ = ar², a₄ = ar³

By the given condition,

a₃ = a₁ + 9

⇒ ar² = a + 9 … (1)

a₂ = a₄ + 18

⇒ ar = ar³ + 18 … (2)

From (1) and (2), we obtain

a(r² ­­– 1) = 9 … (3)

ar (1– r²) = 18 … (4)

Dividing (4) by (3), we obtain

Substituting the value of r in (1), we obtain

4a = a + 9

⇒ 3a = 9

∴ a = 3

Thus, 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.

Question 22:

If the terms of a G.P. are a, b and c, respectively. Prove that

Answer:

Let A be the first term and R be the common ratio of the G.P.

According to the given information,

AR^p–1 = a

AR^q–1 = b

AR^r–1 = c

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

Thus, the given result is proved.

Question 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.

Answer:

The first term of the G.P is a and the last term is b.

Therefore, the G.P. is a, ar, ar², ar³, … ar^n–1, where r is the common ratio.

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.

∴1 + 2 + ……….+ (n – 1)

Thus, the given result is proved.

Question 24:

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

Answer:

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 ^.

Question 25:

If a, b, c and d are in G.P. show that .

Answer:

a, b, c, d are in G.P.

Therefore,

bc = ad … (1)

b² = ac … (2)

c² = bd … (3)

It has to be proved that,

(a² + b² + c²) (b² + c² + d²) = (ab + bc – cd)²

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.

∴ L.H.S. = R.H.S.

∴

Question 26:

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

Answer:

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

Let a be the first term and r be the common ratio of the G.P.

∴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

Thus, the required two numbers are 9 and 27.

Question 27:

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

Answer:

G. M. of a and b is .

By the given condition, 

Squaring both sides, we obtain

Question 28:

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

Answer:

Let the two numbers be a and b.

G.M. = 

According to the given condition,

Also,

Adding (1) and (2), we obtain

Substituting the value of a in (1), we obtain

Thus, the required ratio is.

Question 29:

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

Answer:

It is given that A and G are A.M. and G.M. between two positive numbers. Let these two positive numbers be aand b.

From (1) and (2), we obtain

a + b = 2A … (3)

ab = G² … (4)

Substituting the value of a and b from (3) and (4) in the identity (a – b)² = (a + b)² – 4ab, we obtain

(a – b)² = 4A² – 4G² = 4 (A²–G²)

(a – b)² = 4 (A + G) (A – G)

From (3) and (5), we obtain

Substituting the value of a in (3), we obtain

Thus, the two numbers are.

Question 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:

It is given that the number of bacteria doubles every hour. Therefore, the number of bacteria after every hour will form a G.P.

Here, a = 30 and r = 2

∴ a₃ = ar² = (30) (2)² = 120

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

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

Thus, number of bacteria at the end of n^th hour will be 30(2)ⁿ.

Question 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?

Answer:

The amount deposited in the bank is Rs 500.

At the end of first year, amount = = 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

∴Amount at the end of 10 years = Rs 500 (1.1) (1.1) … (10 times)

= Rs 500(1.1)¹⁰

Question 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 the root of the quadratic equation be a and b.

According to the given condition,

The quadratic equation is given by,

x²– x (Sum of roots) + (Product of roots) = 0

x² – x (a + b) + (ab) = 0

x² – 16x + 25 = 0 [Using (1) and (2)]

Thus, the required quadratic equation is x² – 16x + 25 = 0