Question

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 20${}^{th}$ term is -112.

Answer 1

$T_1+T_2+T_3+T_4+T_5=\frac{1}{4}[T_6+T_7+T_8+T_9+T_{10}]$

Let first term $=a$

common difference$=d$

$[a+(a+d)+(a+2d)+(a+3d)+(a+4d\left)\right]=\frac{1}{4}\left[\right(a+5d)+(a+6d)+(a+7d)+(a+8d)+(a+9d\left)\right]$

$\Rightarrow (5a+10d)=\frac{1}{4}(5a+35d)$

$\Rightarrow 20a+40d=5a+35d$

$\Rightarrow 15a+5d=0$

$\Rightarrow 3a+d=0$

$\Rightarrow d=-3a$

$d=-6$ (given $a=2$)

$T_{20}=a+19d=2+19(-6)$

$=-112$.