For an $A.P.$,the $10^{th}$ term is $52$ and the $16^{th}$ term is $82$. Find the $n^{th}$ term and the $32^{nd}$ term of the $A.P.$

(A) Let the first term be $a$ and the common difference be $d$. The $n^{th}$ term of an $A.P.$ is given by $T_n = a + (n - 1)d$.
Given $T_{10} = 52$,we have $a + 9d = 52$ (Equation $1$).
Given $T_{16} = 82$,we have $a + 15d = 82$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(a + 15d) - (a + 9d) = 82 - 52$,which gives $6d = 30$,so $d = 5$.
Substituting $d = 5$ into Equation $1$: $a + 9(5) = 52$,so $a + 45 = 52$,which gives $a = 7$.
The $n^{th}$ term is $T_n = a + (n - 1)d = 7 + (n - 1)5 = 7 + 5n - 5 = 5n + 2$.
The $32^{nd}$ term is $T_{32} = 5(32) + 2 = 160 + 2 = 162$.