Sequences and Series – exercise 9.2 [11 – 18]- Class XI

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

a/p(q – r)+b/q(r – p)+c/r(p-q) = 0

Solution:

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

According to the given information,

Sp = p/2[2a1 + (p – 1)d] = a

2a1 + (p – 1)d = 2a/p ————(1)

Sq = q/2[2a1 + (q – 1)d] = b

2a1 + (p – 1)d = 2b/p ————(1)

Sr = r/2[2a1 + (r – 1)d] = c

2a1 + (r – 1)d = 2a/r ————(1)

Subtracting (2) from (1), we obtain

(p – 1)d – (q – 1)d = 2a/p2b/q

d( p – 1 – q + 1) = (2aq – 2bq)/pq

d(p – q) = (2aq – 2bq)/pq

d = 2(aq – bq)/[(pq)(p-q)] ——————-(4)

Subtracting (3) from (2), we obtain,

(q – 1)d – (r – 1)d = 2b/q2c/r

d( q – 1 – r + 1) = 2b/q2c/r

d(q – r) = (2br – 2qc)/qr

d = 2(br-qc)/[(qr)(q-r)] ————(5)

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

aq-bp/pq(p-q) = br-qc/qr(q-r)

(aq-bp) qr(q-r) = (br-qc) pq(p-q)

(aq-bp) r(q-r) = (br-qc) p(p-q)

(aqr-bpr) (q-r) = (bpr-qpc) (p-q)

Dividing both the sides by pqr, we obtain

Sequences and Series – exercise 9.2 [11 – 18]- Class XI

Thus, the given result is proved.


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

Solution:

Let a and b be the first term and the common difference of the A.P. respectively. According to the given condition,

Sequences and Series – exercise 9.2 [11 – 18]- Class XI

Thus the given result is proved.


  1. If the sum of n terms of an A.P. is 3n² + 5n and its mth term is 164, find the value of m.

Solution:

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

am = a +(m-1)d = 164 ———(1)

Sum of terms: Sn = n/2[2a+(n-1)d]

Here,

n/2[2a+(n-1)d]=3n2 + 5n

na + n2.d/2 = 3n2 + 5n

Comparing the coefficient of n2 on both sides, we get

d/2 = 3

d = 6

Comparing the coefficient of n on both sides, we get

a – d/2  = 5

a – 3 = 5

a = 8

Therefore from (1)

8 + (m – 1) 6 = 164

⇒ (m – 1) 6 = 164 – 8 = 156

⇒ m – 1 = 26

⇒ m = 27

Thus, the value of m is 27.


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

Solution:

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.


  1. If [an + bn]/[an-1 + bn-1] is the A.M. between a and b then find the value of n.

Solution:

A.M. of a and b = a+b/2

According to the given condition,

Sequences and Series – exercise 9.2 [11 – 18]- Class XI


  1. 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 7th and (m – 1)th numbers is 5:9. Find the value of m.

Solution:

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

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

∴ 31 = 1 + (m + 2 – 1) (d)

⇒ 30 = (m + 1) d

d = 30/m+1 ———-(1)

A1 = a + d

A2 = a + 2d

A3 = a + 3d …

∴ A7 = a + 7d

Am–1 = a + (m – 1) d

According to the given condition,

Sequences and Series – exercise 9.2 [11 – 18]- Class XI

Thus the value of m is 14.


  1. 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 30th installment?

Solution:

The first installment of the loan is Rs 100.

The second instalment 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 30th installment is Rs 245.


  1. 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.

Solution:

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

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

Sn = 180°(n-2)

n/2[2a+(n-1)d] = 180°(n-2)

n/2[240+(n-1)5 °] = 180°(n-2)

n[240+(n-1)5] = 360(n-2)

5n2 +235n – 360n +720 = 0

5n2 – 15n + 720 = 0

n2 – 25n + 144 = 0

n2 – 16n -9n + 144= 0

n(n-16)-9(n-16)=0

(n-16)(n-9)=0

n = 9 or n = 16