Q. 163.9( 118 Votes )

# 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 A_{1}, A_{2}, … A_{m} be m numbers such that 1, A_{1}, A_{2}, … A_{m}, 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

⇒

A_{1} = a + d

A_{2} = a + 2d

A_{3} = a + 3d …

∴ A_{7} = 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.

A_{m-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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