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

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

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

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

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

$A_{1}=a+d$

$A_{2}=a+2 d$

$A_{3}=a+3 d$

$\therefore A_{7}=a+7 d$

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

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 $14$