Question

Between 1 and 31 are inserted m arithmetic means so that the ratio of the 7th and $(m-1)$ the 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 $2m+2=75m-1020\therefore 73m=1022$

$\therefore m=\frac{1022}{73}=14.$