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

Solution:

Let A₁, A₂, ....., Am be numbers such that 1, A₁, A₂, ...., A_m, 31 are in A.P.

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

Therefore,

⇒ 31 = 1+ (m + 2 - 1) d

⇒ 30 = (m + 1) d

d = 30/(m + 1) ....(1)

Hence,

A₁ = a + d

A₂ = a + 2d

A₃ = a + 3d

A₇ = a + 7d

A_{m - 1} = a + (m - 1) d

According to the given condition,

(a + 7d)/[a + (m - 1) d] = 5/9

⇒ 1 + 7 (30/(m + 1))/[1 + (m - 1)30/(m + 1)] = 5/9

⇒ [m + 1 + 7 (30)]/[m + 1 + 30 (m - 1)] = 5/9

⇒ (m + 1 + 210)/(m + 1 + 30m - 30) = 5/9

⇒ (m + 211)/(31m - 29) = 5/9

⇒ 9m + 1899 = 155m - 145

⇒ 155m - 9m = 1899 + 145

⇒ 146m = 2044

⇒ m = 14

Thus, the value of m = 14

---

NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.2 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

Summary:

The value of m came out to be 14 given that the ratio of the 7th and the (m-1)th numbers is 5:9