The $8^{th}$ term of an $A.P.$ is $31$ and its $15^{th}$ term exceeds its $11^{th}$ term by $16$. Find the $A.P.$ and also its $20^{th}$ term.

(N/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 $T_8 = 31$,so $a + 7d = 31$ (Equation $1$).
Given $T_{15} = T_{11} + 16$,so $(a + 14d) = (a + 10d) + 16$.
This simplifies to $4d = 16$,which gives $d = 4$.
Substituting $d = 4$ into Equation $1$: $a + 7(4) = 31 \implies a + 28 = 31 \implies a = 3$.
The $A.P.$ is $a, a+d, a+2d, \ldots$ which is $3, 7, 11, 15, \ldots$.
The $20^{th}$ term is $T_{20} = a + 19d = 3 + 19(4) = 3 + 76 = 79$.