Question

Between $1$ and $31$ are inserted m arithmetic means so that the ratio of the $7$th and $(m-1)$th means is $5:9$. Find the value of m.

Answer 1

Let the means be $x_1,x_2$, _________$x_m$ so that $1$, $x_1,x_2,...x_m$, $31$ is an A.P. of $(m+2)$ terms.
$31=T_{m+2}=a+(m+1)d=1+(m+1)d$
$\therefore d=\frac{30}{m+1}$ ..$\left(1\right)$
We are given that
$\frac{x_7}{x_{m-1}}=\frac{5}{9}$
$\therefore \frac{T_8}{T_m}=\frac{a+7d}{a+(m-1)d}=\frac{5}{9}$
or $9a+63d=5a+(5m-5)d$
or $4.1=(5m-68)\frac{30}{m+1}$ by $\left(1\right)$
or $2m+2=75m-1020$
$\therefore 73m=1022$
$\therefore m=\frac{1022}{73}=14$.