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.

Given:

$a=2, S_{5}=\frac{1}{4}\left(S_{10}-S_{5}\right)$

We have:

$S_{5}=\frac{5}{2}[2 \times 2+(5-1) d]$

$\Rightarrow S_{5}=5[2+2 d] \ldots(i)$

Also, $S_{10}=\frac{10}{2}[2 \times 2+(10-1) d]$

$\Rightarrow S_{10}=5[4+9 d] \ldots . .(i i)$

$\because S_{5}=\frac{1}{4}\left(S_{10}-S_{5}\right)$

From (i) and (ii), we have:

$\Rightarrow 5[2+2 \mathrm{~d}]=\frac{1}{4}[5(4+9 d)-5(2+2 d)]$

$\Rightarrow 8+8 d=4+9 d-2-2 d$

$\Rightarrow d=-6$

$\therefore a_{20}=a+(20-1) d$

$\Rightarrow a_{20}=a+19 d$

$\Rightarrow a_{20}=2+19(-6)$

$\Rightarrow a_{20}=-112$