Question

Between 1 and 31, \(\mathrm{n}\) numbers have been inserted in such a way that the resulting sequence is an A.P. If the ratio of \((\mathrm{n}-1)^{\text{th}}\) and \(7^{\text{th}}\) inserted numbers is \(9: 5\), then \(\mathrm{n}\) is equal to ________.

Hint:

The A.P. has n + 2 terms with first term 1 and last term 31, so the common difference is 30/(n + 1); the k-th inserted number is 1 + k d.

Answer 1

Correct Answer: 14

The A.P. is \(1, a_1, a_2, \dots, a_n, 31\) with \(n+2\) terms, so \(31=1+(n+1)d\) and \(d=\frac{30}{n+1}\).
The k-th inserted number is \(1+kd\).
Given \(\frac{1+(n-1)d}{1+7d}=\frac95\), so \(5+5(n-1)d=9+63d\), i.e. \(d\,(5n-68)=4\).
Substitute \(d=\frac{30}{n+1}\): \(30(5n-68)=4(n+1)\), i.e. \(150n-2040=4n+4\).
\(146n=2044 \Rightarrow n=14\).
Check: \(d=2\); the 13th inserted number is 27, the 7th is 15, and the ratio \(27:15=9:5\).
Answer: 14