NCERT Solutions for Class 11 Maths Chapter 9 – Sequences and Series Ex 9.2

NCERT Solutions for Class 11 Maths Chapter 9 – Sequences and Series Ex 9.2

Page No 185:

Question 1:

Find the sum of odd integers from 1 to 2001.

Answer:

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

This sequence forms an A.P.

Here, first term, a = 1

Common difference, d = 2

Thus, the sum of odd numbers from 1 to 2001 is 1002001.

Question 2:

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

Answer:

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

Thus, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.

Question 3:

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 20^thterm is –112.

Answer:

First term = 2

Let d be the common difference of the A.P.

Therefore, the A.P. is 2, 2 + d, 2 + 2d, 2 + 3d, …

Sum of first five terms = 10 + 10d

Sum of next five terms = 10 + 35d

According to the given condition,

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

Question 4:

How many terms of the A.P. are needed to give the sum –25?

Answer:

Let the sum of n terms of the given A.P. be –25.

It is known that, , where n = number of terms, a = first term, and d = common difference

Here, a = –6

Therefore, we obtain

Question 5:

In an A.P., if p^th term is and q^th term is , prove that the sum of first pq terms is 

Answer:

It is known that the general term of an A.P. is a_n = a + (n – 1)d

∴ According to the given information,

Subtracting (2) from (1), we obtain

Putting the value of d in (1), we obtain

Thus, the sum of first pq terms of the A.P. is .

Question 6:

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

Answer:

Let the sum of n terms of the given A.P. be 116.

Here, a = 25 and d = 22 – 25 = – 3

However, n cannot be equal to . Therefore, n = 8

∴ a₈ = Last term = a + (n – 1)d = 25 + (8 – 1) (– 3)

= 25 + (7) (– 3) = 25 – 21

= 4

Thus, the last term of the A.P. is 4.

Question 7:

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

Answer:

It is given that the k^th term of the A.P. is 5k + 1.

k^th term = a_k = a + (k – 1)d

∴ a + (k – 1)d = 5k + 1

a + kd – d = 5k + 1

Comparing the coefficient of k, we obtain d = 5

a – d = 1

⇒ a – 5 = 1

⇒ a = 6

Question 8:

If the sum of n terms of an A.P. is (pn + qn²), where p and q are constants, find the common difference.

Answer:

It is known that,

According to the given condition,

Comparing the coefficients of n² on both sides, we obtain

∴ d = 2 q

Thus, the common difference of the A.P. is 2q.

Question 9:

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

Answer:

Let a₁, a₂, and d₁, d₂ be the first terms and the common difference of the first and second arithmetic progression respectively.

According to the given condition,

Substituting n = 35 in (1), we obtain

From (2) and (3), we obtain

Thus, the ratio of 18^th term of both the A.P.s is 179: 321.

Question 10:

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

Answer:

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

Here,

According to the given condition,

Thus, the sum of the first (p + q) terms of the A.P. is 0.

Question 11:

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

Prove that 

Answer:

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

According to the given information,

Subtracting (2) from (1), we obtain

p – 1d – q – 1d = 2ap – 2bq⇒dp – 1 – q + 1 = 2aq – 2bppq

⇒dp – q = 2aq – 2bppq

⇒d = 2aq – bppqp – q                …….4

Subtracting (3) from (2), we obtain

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

aq – bppqp – q = br – qcqrq – r

⇒aq – bppp – q = br – qcrq – r

⇒rq – raq – bp = pp – qbr – qc

⇒raq – bpq – r = pbr – qcp – q

⇒aqr – bprq – r = bpr – cpqp – q

Dividing both sides by pqr, we obtain

Thus, the given result is proved.

Question 12:

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

Answer:

Let a and b be the first term and the common difference of the A.P. respectively.

According to the given condition,

Putting m = 2m – 1 and n = 2n – 1 in (1), we obtain

From (2) and (3), we obtain

Thus, the given result is proved.

Question 13:

If the sum of n terms of an A.P. is and its m^th term is 164, find the value of m.

Answer:

Let a and b be the first term and the common difference of the A.P. respectively.

a_m = a + (m – 1)d = 164 … (1)

Sum of n terms,

Here,

n2 2a + nd – d = 3n2 + 5n⇒na + d2n2 – d2n = 3n2 + 5n⇒d2n2 + a – d2n = 3n2 + 5n

Comparing the coefficient of n² on both sides, we obtain

Comparing the coefficient of n on both sides, we obtain

Therefore, from (1), we obtain

8 + (m – 1) 6 = 164

⇒ (m – 1) 6 = 164 – 8 = 156

⇒ m – 1 = 26

⇒ m = 27

Thus, the value of m is 27.

Question 14:

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

Answer:

Let A₁, A₂, A₃, A₄, and A₅ be five numbers between 8 and 26 such that

8, A₁, A₂, A₃, A₄, A₅, 26 is an A.P.

Here, a = 8, b = 26, n = 7

Therefore, 26 = 8 + (7 – 1) d

⇒ 6d = 26 – 8 = 18

⇒ d = 3

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

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

Question 15:

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

Answer:

A.M. of a and b 

According to the given condition,

Question 16:

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.

Answer:

Let A₁, A₂, … A_m be m numbers such that 1, A₁, A₂, … A_m, 31 is an A.P.

Here, a = 1, b = 31, n = m + 2

∴ 31 = 1 + (m + 2 – 1) (d)

⇒ 30 = (m + 1) d

A₁ = a + d

A₂ = a + 2d

A₃ = a + 3d …

∴ A₇ = a + 7d

A_m–1 = a + (m – 1) d

According to the given condition,

Thus, the value of m is 14.

Page No 186:

Question 17:

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

Answer:

The first installment of the loan is Rs 100.

The second installment of the loan is Rs 105 and so on.

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

The A.P. is 100, 105, 110, …

First term, a = 100

Common difference, d = 5

A₃₀ = a + (30 – 1)d

= 100 + (29) (5)

= 100 + 145

= 245

Thus, the amount to be paid in the 30^th installment is Rs 245.

Question 18:

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.

Answer:

The angles of the polygon will form an A.P. with common difference d as 5° and first term a as 120°.

It is known that the sum of all angles of a polygon with n sides is 180° (n – 2).