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