Question

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

Answer 1

Given,

$S_5=\frac{1}{4}(S_{10}-S_5)$

$4S_5=S_{10}-S_5$

$5S_5=S_{10}$...........A

$S_5=\frac{5}{2}\left[2\right(2)+(5-1\left)d\right]$

$=10+10d$

$S_{10}=\frac{10}{2}\left[2\right(2)+(10-1\left)d\right]$

$=20+45d$

From A

$5(10+10d)=20+45d$

$10+10d=4+9d$

$d=-6$

$T_{20}=a+(20-1)d$

$=2+19\times (-6)$

$=-112$

Hence proved.