Question

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

• A. $12$
• B. $13$
• C. $14$
• D. $15$

Answer 1

The correct option is C $14$

Let the arithmetic mean inserted be $a_1,a_2,a_3,...,a_m$.

Now, the series is $1,a_1,a_2,a_3,...,a_m,31$.

Hence $a=1,T_n=31,n=m+2$.

Now, $T_n=a+(n-1)d$

$\Rightarrow 31=1+(m+2-1)d$

$\Rightarrow d=\frac{30}{(m+1)}$ $-----\left(1\right)$

Given $\frac{T_7}{T_{m-1}}=\frac{5}{9}$

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

Here, $a+(7-1+1)d$, $+1$ after $7-1$ is because in the series the first A.M is second in the terms of series.

So, $\frac{a+7d}{a+(m-1)d}=\frac{5}{9}$

$\Rightarrow 9a+63d=5a+5md-5d$

$\Rightarrow 4a=d(5m-68)$

Substitute the values of $d$ and $a$ from equation $\left(1\right)$; we get

$\Rightarrow 4(m+1)=30(5m-68)$

$\Rightarrow 146m=2044$

$\Rightarrow m=14$