The $10^{th}$ term of an $A.P.$ is $52$ and its $17^{th}$ term exceeds its $13^{th}$ term by $20$. Find the $A.P.$ and also its $30^{th}$ term.

(A) Let the first term of the $A.P.$ 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,$T_{10} = 52 \implies a + 9d = 52$ (Equation $1$).
Also,$T_{17} - T_{13} = 20$.
$(a + 16d) - (a + 12d) = 20 \implies 4d = 20 \implies d = 5$.
Substituting $d = 5$ in Equation $1$: $a + 9(5) = 52 \implies a + 45 = 52 \implies a = 7$.
The $A.P.$ is $a, a+d, a+2d, \ldots$ which is $7, 12, 17, \ldots$.
The $30^{th}$ term is $T_{30} = a + 29d = 7 + 29(5) = 7 + 145 = 152$.