In an $A.P.$,the first term is $2$ and the sum of the first five terms is one-fourth of the next five terms. Show that the $20^{th}$ term is $-112$.

(A) Let the first term be $a = 2$ and the common difference be $d$.
The sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
The sum of the first five terms is $S_5 = \frac{5}{2}[2(2) + (5-1)d] = \frac{5}{2}[4 + 4d] = 5(2 + d) = 10 + 5d$.
The sum of the next five terms is the sum of the first ten terms minus the sum of the first five terms: $S_{10} - S_5$.
$S_{10} = \frac{10}{2}[2(2) + (10-1)d] = 5[4 + 9d] = 20 + 45d$.
Sum of the next five terms $= (20 + 45d) - (10 + 5d) = 10 + 40d$.
According to the given condition,$S_5 = \frac{1}{4}(\text{sum of next five terms})$.
$10 + 5d = \frac{1}{4}(10 + 40d)$.
$40 + 20d = 10 + 40d$.
$30 = 20d \Rightarrow d = 1.5$.
Wait,re-evaluating the sum of terms: $S_5 = 2 + (2+d) + (2+2d) + (2+3d) + (2+4d) = 10 + 10d$.
Sum of next five terms $= (2+5d) + (2+6d) + (2+7d) + (2+8d) + (2+9d) = 10 + 35d$.
Given $10 + 10d = \frac{1}{4}(10 + 35d)$.
$40 + 40d = 10 + 35d$.
$5d = -30 \Rightarrow d = -6$.
The $20^{th}$ term is $a_{20} = a + (20-1)d = 2 + 19(-6) = 2 - 114 = -112$.
Thus,the $20^{th}$ term is $-112$.