# 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.

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

Here, First term = a = 1,

Last term = b = 31,

Total no. of terms = 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

It is because here the 7th term is between 1 and 31. So counting from 1 the 7th term will be the 8th term.

Am-1 = a + (m – 1)d

Also given,

9 (m + 211) = 5 (31m – 29)

9m + 1899 = 155m - 145

155m – 9m = 1899 + 145

146m = 2044

m = 14

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