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 the value of $m$ is

• A. $9$
• B. $11$
• C. $13$
• D. $14$

Answer 1

The correct option is D $14$

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: $\frac{x_7}{x_{m-1}}=\frac{5}{9}$

$\therefore \frac{T_8}{T_m}=\frac{a+7d}{a+(m-1)d}=\frac{5}{9}$

$\Rightarrow 9a+63d=5a+(5m-5)d$

$\Rightarrow 4\cdot 1=(5m-68)\frac{30}{m+1}$

$\Rightarrow 2m+2=75m-1020\Rightarrow 73m=1022$

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