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 $7^{th}$ and $(m-1{)}^{th}$ numbers is $5:9$. Then, the value of $m$ is

• A. 10
• B. 14
• C. 12
• D. 15

Answer 1

Answer: (B)

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