Question

The 14^th 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

Here it is given that,

`T_14=2(T_8)`

`a+(14-1)d=2[a+(8-1)d]`

`a+13d=2[a+7d]`

`a+13d=2a+14d`

`13d-14d=2a-a`

-d=a......................(1)

Now, it is given that its 6^th term is –8.

`T_6=-8`

a+(6-1)d=-8

a+5d=-8

-d+5d=-8  [because using (1)]

4d=-8

d=-2

Subs. this in eq. (1), we get a = 2
Now, the sum of 20 terms,

`S_n=n/2[2a+(n-1)d]`

`S_20=20/2[2a+(20-1)d]`

=10[2(2)+19(-2)]

=10[4-38]

-340