Question

How do you find the first three terms of the arithmetic series `a_1=11`, `a_n=110`, `S_n=726`?

Answer 1

First three terms: `11, 20, 29`.

Explanation:

Write a systems of equations.

`t_n = a + (n - 1)d`

`s_n = n/2(2a + (n - 1)d)`

So,

`110 = 11 + (n - 1)d`
`726 = n/2(22 + (n - 1)d)`

Simplify before using substitution.

`726 = n/2(22 + nd - d)`

`726 = (22n + n^2d - nd)/2`

`1452 - 22n = n^2d - nd`

`1452 - 22n = d(n^2 - n)`

`(1452 - 22n)/(n^2 - n) = d`

`110 = 11 + (n - 1)((1452 - 22n)/(n^2 - n))`

`99 = (n - 1)(1452 - 22n)/(n(n - 1))`

`99 = (1452 - 22n)/n`

`99n = 1452 - 22n`

`121n = 1452`

`n = 12`

Now, we can substitute into equation `1` to find the common difference.

`110 = 11 + (12 - 1)d`

`99 = 11d`

`d = 9`

The first 3 terms are `11, 20 and 29`.

Hopefully this helps!