# Thread: Converging geometric sequence with the sum cubed

1. ## Converging geometric sequence with the sum cubed

The sum of the terms of an infinite geometric progression is 4 and the sum of the cubes of the terms is 192. Find the first three terms.

Obviously you can't use the formula for the sum of a convergent series because you don't know the first term or the common ratio...

I have no idea what else to use.

2. Originally Posted by KateMackay
The sum of the terms of an infinite geometric progression is 4 and....
So:

$4\, =\, \frac{a}{1\, -\, r}$

Originally Posted by KateMackay
...the sum of the cubes of the terms is 192.
So:

$192\, =\, \frac{a^3}{1\, -\, r^3}$

Note that:

$1\, -\, r^3\, =\, (1\, -\, r)(1\, +\, r\, +\, r^2)$

3. Okay, so what I've done is used the equation with 4 to solve for r and substitute that value into the equation with 192. Now there is way too much simplifying/factoring for my liking but I'll keep trying.
Thanks!

4. Originally Posted by KateMackay
Okay, so what I've done is used the equation with 4 to solve for r and substitute that value into the equation with 192. Now there is way too much simplifying/factoring for my liking but I'll keep trying.
Thanks!
here's a hint ...

$4 = \frac{a}{1-r}$

$64 = \frac{a^3}{(1-r)^3}$

finally, note that ...

$\frac{192}{64} = 3$