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 $: \frac{x_{7}}{x_{m-1}}=\frac{5}{9}$
$\therefore \frac{T_{8}}{T_{m}}=\frac{a+7 d}{a+(m-1) d}$
$=\frac{5}{9}$
$\Rightarrow 9 a+63 d=5 a+(5 m-5) d$
$\Rightarrow 4.1=(5 m-68) \frac{30}{m+1}$
$\Rightarrow 2 m+2=75 m-1020$
$\Rightarrow 73 m=1022$
$\therefore m=\frac{1022}{73}=14$