# 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 A_{1}, A_{2}, ....., Am be numbers such that 1, A_{1}, A_{2}, ...., 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,

A_{1} = a + d

A_{2} = a + 2d

A_{3} = a + 3d

A_{7} = 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

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