Question

How to use arithmetric progession formula in these question ?

Answer 1

`6^(th)` term to `18^(th)` term are `-12,-15,-18,-21,-24,-27,-30,-33,-36,-39,-42,-45,-48`

Explanation:

We have `S_n=(9n)/2-(3n^2)/2`. Let the `n^(th)` term of A.P. be written as `a_n`.

Now as `S_1=a_1`, but `S_1=9/2-3/2=6/2=3`, hence `a_1=3`

and as `S_2=(9xx2)/2-(3xx2^2)/2=9-6=3`, but `S_2=a_1+a_2`

hence `a_2=S_2-a_1=3-3=0`

Hence common difference is `d=0-3=-3`

Let us check as `S_n=n/2(2a_1+(n-1)d)`

= `n/2(2xx3+(n-1)xx(-3))`

= `n/2(6-3n+3)=n/2(9-3n)=(9n)/2-(3n^2)/2`

and `a_n=3+(n-1)(-3)=6-3n`

Hence `a_6=6-3xx6=-12`

and the sequence from `a_6` to `a_18` is

`-12,-15,-18,-21,-24,-27,-30,-33,-36,-39,-42,-45,-48`