Question

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 $(n-1{)}^{th}$ numbers is $5:9$ . Find the value of $n$.

Answer 1

Let $A_1,A_2...A_n$ be n arithmetic means between $1$ and $31$

Then $1,A_{2,}{A}_{3,}...A_{n,}31$ is an A.P with common difference $d=\frac{31-1}{n+1}=\frac{30}{n+1}$

Now $A_7=1+7d=1+\frac{7\times 30}{n+1}=\frac{n+211}{n+1}$

And $A_{n-1}=1+(n-1)d=1+\frac{30}{n+1}(n-1)=\frac{31n-29}{n+1}$

It is given that

$\frac{A_7}{A_{n-1}}=\frac{5}{9}$

$\Rightarrow \frac{n+211}{31n-29}=\frac{5}{9}$

$\Rightarrow 9n+1899=155n-145$

$\Rightarrow n=14$