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 $112$.

Solution: Here, First term$\left(a_1\right)=2$

If $S_5$ is the sum of first five term and, $S_{10}$ is the sum of first 10 terms, then, sum of 6 terms to 10 terms will be
$S_{10}-S_5$.
We are given,
$S_5=\frac{1}{4}\left(S_{10}-S_5\right)$
$\Rightarrow \frac{5}{4}{S}_5=\frac{1}{4}{S}_{10}$
$\Rightarrow 5S_5=S_{10}$
$\Rightarrow 5\ast \left(\frac{5}{2}\left(2a_1+4d\right)\right)=\left(\frac{10}{2}\left(2a_1+9d\right)\right)$
$\Rightarrow 5a_1+10d=2a_1+9d$
$\Rightarrow d=-3a_1$
As $a_1=2$
$\Rightarrow d=-6$Now, $a_{20}=a_1+\left(20-1\right)\cdot d$
$a_{20}=2+\left(20-1\right)\ast 6=-112$