Question

The sum to infinity if a GP is 16 and the sum of the first 4 terms is 15. Find the first four terms?

Answer 1

The first four terms may either be

`8, 4, 2, 1`

OR

`24, -12, 6, -3`

Explanation:

We know that the sum of an infinite geometric series is

`s_n = a/(1 - r)`

The question tells us that `s_n = 16`.

`16 = a/(1 - r) -> 16(1 - r) = a`

Next we recall that the sum of the first n terms of a geometric progression is

`s_N = (a(1 - r^n))/(1 -r)`

`15 = (a(1 - r^4))/(1 - r)`

We can simplify the equation a little before combining it with the other one.

`15 =(a(1 - r^2)(1 + r^2))/(1 -r)`

`15 = (a(1 + r)(1 - r)(1 + r^2))/(1 - r)`

`15/((1 + r)(r^2 + 1)) = a`

We can now see that

`16(1 - r) = 15/((1 +r)(r^2 + 1))`

`16(1 -r)(r^3 + r^2+ r + 1) = 15`

`16(r^3 + r^2 + r + 1 - r^4 - r^3 - r^2 - r) = 15`

`16 - 16r^4 = 15`

`1 = 16r^4`

`1/16 = r^4`

`r = +- 1/2`

We have two possible situations here.

**When ** `r = 1/2`

`16 = a/(1 - 1/2) -> a = 16(1/2) = 8`

The first four terms here are

`8, 4, 2, 1`

When `r =-1/2`

`16 = a/(3/2) -> a =16(3/2) = 24`

The first four terms here are

`24, -12, 6, -3`

Hopefully this helps!