Question

The sum of the first two terms of a geometric sequence is `8/9` and of the first three terms is `26/27`. What is the first term, common ratio and sum to infinity ?

Answer 1

The first term can be `2/3` with common ratio `1/3` and sum `1`.

The first term can be `32/27` with common ratio `-1/4` and sum `128/135`

These are the only two possibilities.

Explanation:

If the first term is `a` and the common ratio is `r`, then we are given:

`a(1+r) = 8/9`

`a(1+r+r^2) = 26/27`

So:

`13/12 = (26*9)/(27*8) = (color(red)(cancel(color(black)(a)))(1+r+r^2))/(color(red)(cancel(color(black)(a)))(1+r)) = (1+r+r^2)/(1+r)`

Cross multiply to get:

`13(1+r) = 12(1+r+r^2)`

Rearrange to get:

`0 = 12r^2-r-1 = (3r-1)(4r+1)`

So:

`r = 1/3 " "` or `" " r = -1/4`

If `r=1/3` then `a=8/(9(1+r)) = 2/3`

The sum of the whole series is then `(2/3) / (1-1/3) = 1`

If `r=-1/4` then `a=8/(9(1+r)) = 32/27`

The sum of the whole series is then `(32/27) / (1-(-1/4)) = 128/135`