Search any question & find its solution
Question:
Answered & Verified by Expert
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 $7^{\text {th }}$ and $(m-1)^{\text {th }}$ numbers is $5: 9$. Find the value of $m$.
Solution:
1593 Upvotes
Verified Answer
Let $A_1, A_2, \ldots \ldots \ldots A_m$ be the $\mathrm{m}$ arithmetic means each inserted between 1 and 31 .
Total number of terms $=m+2$
If $d$ is the common difference, then
$\begin{aligned}
&\Rightarrow T_{m+2}=1+(m+2-1) d=31 \Rightarrow(m+1) d \\
&=30 \Rightarrow d=\frac{30}{m+1} \\
&A_7=T_8=a+7 d
\end{aligned}$
$=1+7 \times \frac{30}{m+1}=\frac{m+1+210}{m+1}=\frac{m+211}{m+1}$
$A_{m-1}=T_m$
$=a+(m-1) d$
$=1+(m-1) \times \frac{30}{m+1}=\frac{m+1+30 m-30}{m+1}$
$=\frac{31 m-29}{m+1}$
Now, $\quad \frac{A_7}{A_{m-1}}=\frac{m+211}{31 m-29}=\frac{5}{9}$
$\begin{aligned}
&\Rightarrow 9 m+1899=155 m-145 \Rightarrow 146 m=2044 \\
&\Rightarrow m=\frac{2044}{146}=14 \Rightarrow m=14
\end{aligned}$
Total number of terms $=m+2$
If $d$ is the common difference, then
$\begin{aligned}
&\Rightarrow T_{m+2}=1+(m+2-1) d=31 \Rightarrow(m+1) d \\
&=30 \Rightarrow d=\frac{30}{m+1} \\
&A_7=T_8=a+7 d
\end{aligned}$
$=1+7 \times \frac{30}{m+1}=\frac{m+1+210}{m+1}=\frac{m+211}{m+1}$
$A_{m-1}=T_m$
$=a+(m-1) d$
$=1+(m-1) \times \frac{30}{m+1}=\frac{m+1+30 m-30}{m+1}$
$=\frac{31 m-29}{m+1}$
Now, $\quad \frac{A_7}{A_{m-1}}=\frac{m+211}{31 m-29}=\frac{5}{9}$
$\begin{aligned}
&\Rightarrow 9 m+1899=155 m-145 \Rightarrow 146 m=2044 \\
&\Rightarrow m=\frac{2044}{146}=14 \Rightarrow m=14
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.