Question

There are m arithmetic means between $1$ and $31$ such that the ratio of the $7^{th}$ mean to the $(m-1)th$ mean is $5:9$ Find m

Answer 1

We know that to insert $n$ numbers between $a$ and $b$ common difference$d=\frac{b-a}{n+1}$

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

Now, $a=1,d=\frac{30}{m+1},b=31$

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

$\Rightarrow 9+63d=5+5d(m-1)$

$\Rightarrow 4+68d=5dm$

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

$4+68\left(\frac{30}{m+1}\right)=5\left(\frac{30}{m+1}\right)m$

$\Rightarrow \frac{4m+4+68\times 30}{m+1}=\frac{5\times 30\times m}{m+1}$

$\Rightarrow 4(m+1)+2040=150m$

$\Rightarrow 2044=150m-4m$

$\Rightarrow m=\frac{2044}{146}=14$

$\therefore m=14$