Question

Between 1 and $31, m$ numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of $7^{t h}$ and $(m-1)^{t h}$ numbers is $5: 9$. Then, the value of $m$ is

• A. 10
• B. 14
• C. 12
• D. 15

Answer 1

Answer: (B)

Let $A_1, A_2, A_3, \ldots, A_m$ are $m$ numbers inserted between $1$ and $31$.
$\therefore$ We have $1, A_1, A_2, A_3, \ldots \ldots, A_m, 31$ are in A.P.
Now, $ T_n=a+(n-1) d$
$31=1+(m+2-1) d (\because n=m+2 $, as there are $ m $ numbers between $ 1$ and $ 31)$
$\Rightarrow 31-1 =(m+1) d $
$\Rightarrow 30 =(m+1) d$
$\Rightarrow d =\frac{30}{m+1} .....$(i)
Given, $ \frac{A_7}{A_{m-1}}=\frac{T_8}{T_m}=\frac{5}{9}$
$ \Rightarrow \frac{a+7 d}{a+(m-1) d}=\frac{5}{9}$
$ \Rightarrow \frac{1+7 \times \frac{30}{m+1}}{1+(m-1) \times \frac{30}{m+1}}=\frac{5}{9}$ [from Eq. (i)]
$\Rightarrow \frac{\frac{m+1+210}{m+1}}{\frac{(m+1)+30(m-1)}{m+1}}=\frac{5}{9}$
$\Rightarrow \frac{m+211}{m+1+30 m-30}=\frac{5}{9}$
$\Rightarrow 9(m+211)=5(31 m-29)$
$\Rightarrow 9 m+1899=155 m-145$
$\Rightarrow 1899+145=155 m-9 m$
$\Rightarrow 146 m=2044 \Rightarrow m=\frac{2044}{146}=14$