NCERT Solutions Maths Chapter 9 Sequences and Series for Class 11

NCERT Solutions Exercise 9.1 , 9.2 , 9.3, 9.4 Sequences and Series for Class 11

Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

Q1. a_n = n (n + 2)

Given,

n^th term of a sequence a_n = n (n + 2)

On substituting n = 1, 2, 3, 4, and 5, we get the first five terms

a₁ = 1(1 + 2) = 3

a₂ = 2(2 + 2) = 8

a₃ = 3(3 + 2) = 15

a₄ = 4(4 + 2) = 24

a₅ = 5(5 + 2) = 35

Hence, the required terms are 3, 8, 15, 24, and 35.

Q2. a_n = n/n+1

Given the n^th term, a_n = n/n+1

On substituting n = 1, 2, 3, 4, 5, we get

Q3. a_n = 2ⁿ

Given the n^th term, a_n = 2ⁿ

On substituting n = 1, 2, 3, 4, 5, we get

a₁ = 2¹ = 2

a₂ = 2² = 4

a₃ = 2³ = 8

a₄ = 2⁴ = 16

a₅ = 2⁵ = 32

Q4. a_n = (2n – 3)/6

Given the n^th term, a_n = (2n – 3)/6

On substituting n = 1, 2, 3, 4, 5, we get

Q5. a_n = (-1)^n-1 5ⁿ⁺¹

Given the n^th term, a_n = (-1)^n-1 5ⁿ⁺¹

On substituting n = 1, 2, 3, 4, 5, we get

Q6.

On substituting n = 1, 2, 3, 4, 5, we get the first 5 terms.

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose n^th terms are:

Q7. a_n = 4n – 3; a₁₇, a₂₄

Given,

The n^th term of the sequence is a_n = 4n – 3

On substituting n = 17, we get

a₁₇ = 4(17) – 3 = 68 – 3 = 65

Next, on substituting n = 24, we get

a₂₄ = 4(24) – 3 = 96 – 3 = 93

Q8. a_n = n²/2ⁿ ; a⁷

Given,

The n^th term of the sequence is a_n = n²/2ⁿ

Now, on substituting n = 7, we get

a₇ = 7²/2⁷ = 49/ 128

Q9. a_n = (-1)^n-1 n³; a₉

Given,

The n^th term of the sequence is a_n = (-1)^n-1 n³

On substituting n = 9, we get

a₉ = (-1)^9-1 (9)³ = 1 x 729 = 729

Q10.

On substituting n = 20, we get

Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:

Q11. a₁ = 3, a_n = 3a_n-1 + 2 for all n > 1

Solution:

Given, a_n = 3a_n-1 + 2 and a₁ = 3

Then,

a₂ = 3a₁ + 2 = 3(3) + 2 = 11

a₃ = 3a₂ + 2 = 3(11) + 2 = 35

a₄ = 3a₃ + 2 = 3(35) + 2 = 107

a₅ = 3a₄ + 2 = 3(107) + 2 = 323

Thus, the first 5 terms of the sequence are 3, 11, 35, 107 and 323.

Hence, the corresponding series is

3 + 11 + 35 + 107 + 323 …….

Q12. a₁ = -1, a_n = a_n-1/n, n ≥ 2

Given,

a_n = a_n-1/n and a₁ = -1

Then,

a₂ = a₁/2 = -1/2

a₃ = a₂/3 = -1/6

a₄ = a₃/4 = -1/24

a₅ = a₄/5 = -1/120

Thus, the first 5 terms of the sequence are -1, -1/2, -1/6, -1/24 and -1/120.

Hence, the corresponding series is

-1 + (-1/2) + (-1/6) + (-1/24) + (-1/120) + …….

Q13. a₁ = a₂= 2, a_n = a_n-1 – 1, n > 2

Given,

a₁ = a₂, a_n = a_n-1 – 1

Then,

a₃ = a₂ – 1 = 2 – 1 = 1

a₄ = a₃ – 1 = 1 – 1 = 0

a₅ = a₄ – 1 = 0 – 1 = -1

Thus, the first 5 terms of the sequence are 2, 2, 1, 0 and -1.

The corresponding series is

2 + 2 + 1 + 0 + (-1) + ……

Q14. The Fibonacci sequence is defined by

1 = a₁ = a₂ and a_n = a_{n – 1}+ a_{n – 2}, n > 2

Find a_n+1/a_n, for n = 1, 2, 3, 4, 5

Given,

1 = a₁ = a₂

a_n = a_{n – 1}+ a_{n – 2}, n > 2

So,

a₃ = a₂ + a₁ = 1 + 1 = 2

a₄ = a₃ + a₂ = 2 + 1 = 3

a₅ = a₄ + a₃ = 3 + 2 = 5

a₆ = a₅ + a₄ = 5 + 3 = 8

Thus,

Q1. Find the sum of odd integers from 1 to 2001.

Solution:

The odd integers from 1 to 2001 are 1, 3, 5, …1999, 2001.

It clearly forms a sequence in A.P.

Where the first term, a = 1

The common difference, d = 2

Now,

a + (n -1)d = 2001

1 + (n-1)(2) = 2001

2n – 2 = 2000

2n = 2000 + 2 = 2002

n = 1001

We know,

S_n = n/2 [2a + (n-1)d]

Q2. Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995.

It clearly forms a sequence in A.P.

Where the first term, a = 105

The common difference, d = 5

Now,

a + (n -1)d = 995

105 + (n – 1)(5) = 995

105 + 5n – 5 = 995

5n = 995 – 105 + 5 = 895

n = 895/5

n = 179

We know,

S_n = n/2 [2a + (n-1)d]

Q3. In an A.P, the first term is 2, and the sum of the first five terms is one-fourth of the next five terms. Show that the 20^th term is –112.

Given,

The first term (a) of an A.P = 2

Let’s assume d is the common difference of the A.P.

So, the A.P. will be 2, 2 + d, 2 + 2d, 2 + 3d, …

Then,

Sum of first five terms = 10 + 10d

Sum of next five terms = 10 + 35d

From the question, we have

10 + 10d = ¼ (10 + 35d)

40 + 40d = 10 + 35d

30 = -5d

d = -6

a₂₀ = a + (20 – 1)d = 2 + (19) (-6) = 2 – 114 = -112

Therefore, the 20^th term of the A.P. is –112.

Q4. How many terms of the A.P. -6, -11/2, -5, …. are needed to give the sum –25?

Let’s consider the sum of n terms of the given A.P. as –25.

We known that,

S_n = n/2 [2a + (n-1)d]

where n = number of terms, a = first term, and d = common difference

So here, a = –6

d = -11/2 + 6 = (-11 + 12)/ 2 = 1/2

Thus, we have

Q5. In an A.P., if p^th term is 1/q and q^th term is 1/p, prove that the sum of first pq terms is ½ (pq + 1) where p ≠ q.

Q6. If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term.

Solution:

Given A.P.,

25, 22, 19, …

Here,

First term, a = 25 and

Common difference, d = 22 – 25 = -3

Also given, the sum of a certain number of terms of the A.P. is 116.

The number of terms is n.

So, we have

S_n = n/2 [2a + (n-1)d] = 116

116 = n/2 [2(25) + (n-1)(-3)]

116 x 2 = n [50 – 3n + 3]

232 = n [53 – 3n]

232 = 53n – 3n²

3n² – 53n + 232 = 0

3n² – 24n – 29n+ 232 = 0

3n(n – 8) – 29(n – 8) = 0

(3n – 29) (n – 8) = 0

Hence,

n = 29/3 or n = 8

As n can only be an integral value, n = 8

Thus, the 8^th term is the last term of the A.P.

a₈ = 25 + (8 – 1)(-3)

= 25 – 21

= 4

Q7. Find the sum to n terms of the A.P., whose k^thterm is 5k + 1.

Solution:

Given, the k^th term of the A.P. is 5k + 1.

k^th term = a_k = a + (k – 1)d

And,

a + (k – 1)d = 5k + 1

a + kd – d = 5k + 1

On comparing the coefficient of k, we get d = 5

a – d = 1

a – 5 = 1

⇒ a = 6

Q8.

Q9. The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their 18th terms.

Q10. If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find sum of the first (p + q) terms.

Q11. Sum of the first p, q and r terms of an A.P. are a, b and c, respectively.

Prove that

Let a₁ and d be the first term and the common difference of the A.P., respectively.

Then, according to the question, we have

Now, subtracting (2) from (1), we get

Q12. The ratio of the sums of m and n terms of an A.P. is m²: n². Show that the ratio of the m^thand the n^th term is (2m – 1): (2n – 1).

Let’s consider that a and b are the first term and the common difference of the A.P., respectively.

Then, from the question, we have

Hence, the given result is proved.

Q13. If the sum of n terms of an A.P. is 3n² + 5n and its m^thterm is 164, find the value of m.

Let’s consider a and b to be the first term and the common difference of the A.P., respectively.

a_m = a + (m – 1)d = 164 … (1)

The sum of the terms is given by,

S_n = n/2 [2a + (n-1)d]

Q14. Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

Let’s assume A₁, A₂, A₃, A₄, and A₅ to be five numbers between 8 and 26 such that 8, A₁, A₂, A₃, A₄, A₅, 26 are in an A.P.

Here, we have,

a = 8, b = 26, n = 7

So,

26 = 8 + (7 – 1) d

6d = 26 – 8 = 18

d = 3

Now,

A₁ = a + d = 8 + 3 = 11

A₂ = a + 2d = 8 + 2 × 3 = 8 + 6 = 14

A₃ = a + 3d = 8 + 3 × 3 = 8 + 9 = 17

A₄ = a + 4d = 8 + 4 × 3 = 8 + 12 = 20

A₅ = a + 5d = 8 + 5 × 3 = 8 + 15 = 23

Therefore, the required five numbers between 8 and 26 are 11, 14, 17, 20, and 23.

Q15. If is the A.M. between a and b, then find the value of n.

The A.M between a and b is given by (a + b)/2

Then, according to the question,

Q16. Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7^th and (m – 1)^th numbers is 5: 9. Find the value of m.

Let’s consider a₁, a₂, … a_m be m numbers such that 1, a₁, a₂, … a_m, 31 is an A.P.

And here,

a = 1, b = 31, n = m + 2

So, 31 = 1 + (m + 2 – 1) (d)

30 = (m + 1) d

d = 30/ (m + 1) ……. (1)

Now,

a₁ = a + d

a₂ = a + 2d

a₃ = a + 3d …

Hence, a₇ = a + 7d

a_m_–1 = a + (m – 1) d

According to the question, we have

Q17. A man starts repaying a loan as the first instalment of Rs. 100. If he increases the instalment by Rs 5 every month, what amount will he pay in the 30^th instalment?

Given,

The first instalment of the loan is Rs 100.

The second instalment of the loan is Rs 105, and so on as the instalment increases by Rs 5 every month.

Thus, the amount that the man repays every month forms an A.P.

And then, A.P. is 100, 105, 110, …

Where the first term, a = 100

Common difference, d = 5

So, the 30^th term in this A.P. will be

A₃₀ = a + (30 – 1)d

= 100 + (29) (5)

= 100 + 145

= 245

Therefore, the amount to be paid in the 30^th instalment will be Rs 245.

Q18. The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of the sides of the polygon.

It’s understood from the question that the angles of the polygon will form an A.P. with a common difference d = 5° and first term a = 120°.

And we know that the sum of all angles of a polygon with n sides is 180° (n – 2).

Thus, we can say

Q1. Find the 20^th and n^thterms of the G.P. 5/2, 5/4, 5/8, ………

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

Q2. Find the 12^th term of a G.P. whose 8^th term is 192, and the common ratio is 2.

Given,

The common ratio of the G.P., r = 2

And, let a be 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)

Q3. The 5^th, 8^th and 11^th terms of a G.P. are p, q and s, respectively. Show that q² = ps.

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

Q4. The 4^th term of a G.P. is the square of its second term, and the first term is –3. Determine its 7^th term.

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 = arⁿ^–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.

Q5. 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 ?

(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

Q6. 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

Q7. Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015 …

Given G.P., 0.15, 0.015, 0.00015, …

Here, a = 0.15 and r = 0.015/0.15 = 0.1

Q8. Find the sum to n terms in the geometric progression √7, √21, 3√7, ….

The given G.P. is √7, √21, 3√7, ….

Here,

a = √7 and

Q9. Find the sum to n terms in the geometric progression 1, -a, a², -a³ …. (if a ≠ -1)

The given G.P. is 1, -a, a², -a³ ….

Here, the first term = a₁ = 1

And the common ratio = r = – a

We know that,

Q10. Find the sum to n terms in the geometric progression x³, x⁵, x⁷, … (if x ≠ ±1 )

Given G.P. is x³, x⁵, x⁷, …

Here, we have a = x³ and r = x⁵/x³ = x²

Q11. Evaluate:

Q12. 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.

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.

Q13. How many terms of G.P. 3, 3², 3³, … are needed to give the sum 120?

Given G.P. is 3, 3², 3³, …

Let’s consider that n terms of this G.P. be required to obtain the sum 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 120.

Q14. 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.
sol:
Let a be the first term and r be the common ratio, then

Q15. Given a G.P. with a = 729 and 7th term 64, determine S₇.

a = 729, a₇ = 64
Let r be the common ratio of the G.P.
It is known that, a_n = ar^{n – 1}
a₇ = ar^{7 – 1} (729)r⁶

Q16.

Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.
Ans:
Let, first term = a and common ratio = r.
Given, sum of first two terms = – 4
⇒ a₁ + a₂ = – 4
⇒ a + ar = – 4
⇒ a (1 + r) = – 4
and fifth term = 4 × third term
a₅ = 4 × a₃
⇒ ar⁴ = 4ar²
r² = 4
r = ± 2
When r = 2, then from eq. (i), we get
a (1 + 2) = – 4
⇒ a = – $\frac{4}{3}$
Then, G.P. is – $\frac{4}{3}$ , – $\frac{4}{3}$ × 2, – $\frac{4}{3}$ × (2)², ……………
i.e., $\frac{- 4}{3}, \frac{- 8}{3}, \frac{- 16}{3}$
When r = – 2, then from eq. (i), we get a
a (1 – 2)= – 4
⇒ – a = – 4
⇒ a = 4
Then, G.P. is 4, 4 × (- 2), 4 × (- 2)² … i.e, 4,- 8, 16, ………..

Question 17:
If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.
Ans:
Given, 4 th term,
T₄ = x
⇒ ar^{4 – 1} = x
⇒ ar³ = x …………………..(i)
10th term,
T₁₀ = y
⇒ ar^{10 – 1} = y
⇒ ar⁹ = y ………………….(ii)
and 16 th term, T₁₆ = z
⇒ ar^{16 – 1} = z
⇒ ar¹⁵ = z ………………(iii)
Now, multiplying eq. (j) by eq. (ill), we get
ar³ × ar¹⁵ = x × z
a² r^{3 + 15} = x × z
a² r¹⁸ = xz
(ar⁹)² = xz
∴ y² = xz [from eq. (ii)]
Therefore, x,y and z are in GP.

Q18.

Q19.

Q20.

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.

It has to be proved that the sequence, aA, arAR, ar²AR², ………. ar^{n – 1} AR^{n – 1}, forms a G.P.
$\frac{Second term}{First term} = \frac{a r A R}{a A}$ = rR

$\frac{Third term}{Second term} = \frac{a r^{2} A R^{2}}{a r A R}$ = rR
Thus, the above sequence forms a G.P. and the common ratio is rR.

Q21 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.
Ans:
Let a be the first term and r the common ratio of G.P.
∴ nth term = T_n = ar^{n – 1}
⇒ T₂ = ar , T₃ = ar² and T₄ = ar³
Since third term is greater than the first by 9.
∴ T₃ = T₁ + 9
⇒ ar² = a + 9
Second term is greater than the 4th by 18.
T₂ = T₄ + 18
⇒ ar = ar³ + 18
⇒ ar³ = ar + 9r
From eqs. (ii) and (iii), we get
ar = ar + 9r + 18
⇒ 0 = 9r + 18
⇒ r = $\frac{- 18}{9}$ = – 2
Put = – 2 in (i), we get
a(- 2)² = a + 9
⇒ 4a = a + 9
⇒ 3a = 9
⇒ a = 3
T₂ = ar = 3 (- 2) = – 6
T₃ = ar²
= 3 (- 2)² = 12
T₄ = ar³
= 3 (- 2)³ = – 24
∴ Required terms are 3, – 6, 12 and – 24.

Q22.

Q23. If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P² = (ab)ⁿ.

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} …………..(i)
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)}
Here, 1, 2, ………. (n – 1) is an A.P.

Q24.

Let a be the first term and r be the common ratio of the G.P.
Sum of first n terms = a(1−rⁿ)/ (1−r)
Since there are n terms from (n + 1)th to (2n)th term,
sum of terms from (n + 1)th to (2n)th term
= an+1(1−rⁿ)/ (1−r)
= arⁿ(1−rⁿ)/ (1−r) [∵ a_{n + 1} = ar^{n + 1 – 1} = arⁿ]

Thus, required ratio = a(1−rⁿ) / (1−r)×(1−r)arⁿ/ (1−rⁿ)=1/rⁿ

Thus, the ratio of the sum of first n terms of G.P. to the sum of terms from (n + 1)th to (2n)th term is 1/rⁿ

NCERT Solutions Exercise 9.1 , 9.2 , 9.3, 9.4 Sequences and Series for Class 11

Let a be the first term and r be the common ratio of the G.P.
Sum of first n terms = a(1−rⁿ)/ (1−r)
Since there are n terms from (n + 1)th to (2n)th term,
sum of terms from (n + 1)th to (2n)th term
= an+1(1−rⁿ)/ (1−r)
= arⁿ(1−rⁿ)/ (1−r) [∵ a_{n + 1} = ar^{n + 1 – 1} = arⁿ]

Class 11 Maths Chapter 9 Exercise 9.4 pdf

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NCERT Solutions Exercise 9.1 , 9.2 , 9.3, 9.4 Sequences and Series for Class 11

Let a be the first term and r be the common ratio of the G.P.Sum of first n terms = a(1−rn)/ (1−r)Since there are n terms from (n + 1)th to (2n)th term,sum of terms from (n + 1)th to (2n)th term= an+1(1−rn)/ (1−r)= arn(1−rn)/ (1−r) [∵ an + 1 = arn + 1 – 1 = arn]

Class 11 Maths Chapter 9 Exercise 9.4 pdf

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© 2025 Edu-Spot : All Rights Reserved

Let a be the first term and r be the common ratio of the G.P.
Sum of first n terms = a(1−rⁿ)/ (1−r)
Since there are n terms from (n + 1)th to (2n)th term,
sum of terms from (n + 1)th to (2n)th term
= an+1(1−rⁿ)/ (1−r)
= arⁿ(1−rⁿ)/ (1−r) [∵ a_{n + 1} = ar^{n + 1 – 1} = arⁿ]