Question

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

Answer 1

Let $A_a,A_2,A_3.....A_m$ be $m$ numbers between $1$ and $31$ such that $1,A_1,A_2,A_3...A_m,$ $31$ are in A.P. Here $a=1$ and let the common difference be $d$.

Total number of terms in AP $n=m+2$ terms

$\therefore a_{m+2}=31$

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

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

$\Rightarrow d=\frac{31-1}{m+1}$

$\Rightarrow d=\frac{30}{m+1}$

Now, $A_7$= 7th inserted number $=a+7d$

= $1+7\times \left(\frac{30}{m+1}\right)$

= $\frac{m+1+210}{m+1}=\frac{m+211}{m+1}$

$A_{m-1}=a+(m-1)d$

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

= $\frac{m+30m-30}{m+1}=\frac{31m-29}{m+1}$

$\therefore \frac{A_7}{A_{m-1}}=\frac{5}{9}\left[given\right]$

$\therefore \frac{\frac{m+211}{m+1}}{\frac{31m-29}{m+1}}=\frac{5}{9}\Rightarrow \frac{m+211}{31m-29}=\frac{5}{9}$

$\Rightarrow 9m+1899=155m-145$

$\Rightarrow 146m=2044$

$\Rightarrow m=\frac{2044}{146}$

$\Rightarrow m=14.$