Question

If $m$ arithmetic means are inserted between $1$ and $31$ so that the ratio of the $7^{th}$ and $(m-1{)}^{th}$ means is $5:9$, then find the value of m.

• A. 14
• B. 24
• C. 10
• D. 20

Answer 1

Answer: (A)

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.
Now, $31=T_{m+2}=a+(m+1)d=1+(m+1)d$
$\therefore d=\frac{30}{m+1}$
Given $:\frac{x_7}{x_{m-1}}=\frac{5}{9}$
$\therefore \frac{T_8}{T_m}=\frac{a+7d}{a+(m-1)d}$
$=\frac{5}{9}$
$\Rightarrow 9a+63d=5a+(5m-5)d$
$\Rightarrow 4.1=(5m-68)\frac{30}{m+1}$
$\Rightarrow 2m+2=75m-1020$
$\Rightarrow 73m=1022$
$\therefore m=\frac{1022}{73}=14$