Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P.

Solution:

Let $\mathrm{A}_{1}, \mathrm{~A}_{2}, \ldots \mathrm{A}_{m}$ be $m$ numbers such that $1, \mathrm{~A}_{1}, \mathrm{~A}_{2}, \ldots \mathrm{A}_{m}, 31$ is an $\mathrm{A} . \mathrm{P}$.

Here, $a=1, b=31, n=m+2$

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

$\Rightarrow 30=(m+1) d$

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

$\mathrm{A}_{1}=a+d$

$\mathrm{A}_{2}=\mathrm{a}+2 \mathrm{~d}$

$\mathrm{A}_{3}=\mathrm{a}+3 \mathrm{~d} \ldots$

$\therefore \mathrm{A}_{7}=\mathrm{a}+7 d$

$\mathrm{A}_{m-1}=\mathrm{a}+(m-1) d$

According to the given condition,

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

$\Rightarrow \frac{1+7\left(\frac{30}{(m+1)}\right)}{1+(m-1)\left(\frac{30}{m+1}\right)}=\frac{5}{9}$ [From (1)]

$\Rightarrow \frac{m+1+7(30)}{m+1+30(m-1)}=\frac{5}{9}$

$\Rightarrow \frac{m+1+210}{m+1+30 m-30}=\frac{5}{9}$

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

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

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

$\Rightarrow 146 m=2044$

$\Rightarrow m=14$

Thus, the value of m is 14.