The mean of the following frequency distribution is $65$ and the total frequency is $100$. Find the missing frequencies $f_{1}$ and $f_{2}$.

Class	$15-35$	$35-55$	$55-75$	$75-95$	$95-115$
Frequency	$17$	$f_1$	$32$	$f_2$	$19$

(A) Given total frequency $\Sigma f_i = 100$.
Sum of frequencies: $17 + f_1 + 32 + f_2 + 19 = 68 + f_1 + f_2 = 100 \implies f_1 + f_2 = 32$ (Equation $1$).
Using the step-deviation method: $\bar{x} = A + \frac{\Sigma f_i u_i}{\Sigma f_i} \times c$,where $A = 65$ (assumed mean),$c = 20$ (class size).

Class	Frequency $(f_i)$	Midpoint $(x_i)$	$u_i = \frac{x_i - 65}{20}$	$f_i u_i$
$15-35$	$17$	$25$	$-2$	$-34$
$35-55$	$f_1$	$45$	$-1$	$-f_1$
$55-75$	$32$	$65$	$0$	$0$
$75-95$	$f_2$	$85$	$1$	$f_2$
$95-115$	$19$	$105$	$2$	$38$
Total	$100$	-	-	$f_2 - f_1 + 4$

Mean $\bar{x} = 65 + \frac{f_2 - f_1 + 4}{100} \times 20 = 65$.
$65 + \frac{f_2 - f_1 + 4}{5} = 65 \implies f_2 - f_1 + 4 = 0 \implies f_1 - f_2 = 4$ (Equation $2$).
Adding $(1)$ and $(2)$: $2f_1 = 36 \implies f_1 = 18$.
Substituting $f_1 = 18$ in $(1)$: $18 + f_2 = 32 \implies f_2 = 14$.
Thus,$f_1 = 18$ and $f_2 = 14$.