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 20th term is –112.

First term = 2

Let d be the common difference of the A.P.

Therefore, the A.P. is 2, 2 + d, 2 + 2d, 2 + 3d, …

Sum of first five terms = 10 + 10d

Sum of next five terms = 10 + 35d

According to the given condition,

$10+10 d=\frac{1}{4}(10+35 d)$

$\Rightarrow 40+40 d=10+35 d$

$\Rightarrow 30=-5 d$

$\Rightarrow d=-6$

$\therefore a_{20}=a+(20-1) d=2+(19)(-6)=2-114=-112$

Thus, the $20^{\text {th }}$ term of the A.P. is $-112$.