Question

How do you find the first three terms of the arithmetic series n=31, `a_n=78`, `S_n=1023`?

Answer 1

First three terms are `-12`, `-9` and `-6`

Explanation:

In an arithmetic series with `a_1` as first term and `d` as common difference, while `n^(th)` term `a_n=a_1+(n-1)xxd`, the sum of series up to `n^(th)` terms is `S_n=n/2(a_1+a_n)=n/2(2a+(n-1)d)`.

Here we have `n=31`, `a_n=a_1+30d=78`

and `S_n=1023=31/2(a_1+78)`

Hence `a_1=1023xx2/31-78=33xx2-78=-12`

and as `78=-12+30xxd` we have `d=(78+12)/30=90/30=3`

Hence, first three terms are `-12`, `-12+3=-9` and `-9+3=-6`