Question

The 14th term of an A.P. is twice its $8^{th}$ term. If its $6^{th}$ term is $-8$, then find the sum of its first $20$ terms.

Answer 1

Let the first term is $a$ and common difference is $d$

$a_{14}=a+(14-1)d=a+13d$

$a_8=a+(8-1)d=a+7d$

$a_6=a+(6-1)d=a+5d$

Given: $a_{14}$ is twice of $a_8$

$\therefore a+13d=2(a+7d)$

$\Rightarrow a+13d=2a+14d$

$\Rightarrow 2a+14d-a-13d=0$

$\Rightarrow a+d=0$

$\Rightarrow a=-d$ .....(1)

$a_6=a+(6-1)d=a+5d$

$\therefore a+5d=-8$ ......(2)

Put value $a=-d$ in equation (2) we get

$-d+5d=-8$

$\Rightarrow 4d=-8$

$\Rightarrow d=-2$

Substituting the value of $d=-2$ in equation (1) we get

$a=2$

Then $S_{20}=\frac{20}{2}[4+(10-1\left)\right(-2\left)\right]=10(4-38)=-340$