Question

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

Answer 1

Given,

$T_6=-8$

$a+(6-1)d=-8$

$a+5d=-8$

$a=-8-5d$

$T_{14}=2T_8$

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

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

$a=-d$

$-8-5d=-d$

$-8=4d$

$d=-2$

$a=2$

$s_{20}=\frac{20}{10}(2+2+(20-1\left)\right(-2\left)\right)$

$=10(4+19(-2\left)\right)$

$=10(4-38)$

$=-340$