Question

How to calculate these ?

In an arithmetic progression, the sum of the first n term is given by `S_n = (9n)/2 - (3n^2)/2`. Look up the 6th term to the 18th term.

Answer 1

Given sum of first n terms of A P `S_n = (9n)/2 - (3n^2)/2`.

We can write n th term of the series is

`t_n=S_n-S_(n-1)`

`=>t_n =(9n)/2 - (3n^2)/2 -(9(n-1))/2 +(3(n-1)^2)/2`

`=>t_n =(9n)/2 -(9(n-1))/2 - (3n^2)/2 +(3(n-1)^2)/2`

`=>t_n =9/2(n-n+1) - 3/2(n^2 -(n-1)^2)`

`=>t_n =9/2 - 3/2(2n-1)`

So
Inserting `n=6`,

`t_6 =9/2 - 3/2(2xx6-1)=-12`

Inserting `n=7`,

`t_7 =9/2 - 3/2(2xx7-1)=-15`

Inserting `n=8`,

`t_8 =9/2 - 3/2(2xx8-1)=-18`

`---------`
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Inserting `n=18`,

`t_18 =9/2 - 3/2(2xx18-1)=-48`