# Sum of Geometric Progression

• May 31st 2011, 03:20 AM
Blizzardy
Sum of Geometric Progression
Hi guys, need some help with this question:

The sum of the first 20 terms of a geometric series is 10 and the sum of the first 30 terms is 91, find the sum of the first 10 terms.

So this is how i did it:

I tried to find the common ratio:
S20 = 10
a(r^20 - 1)/(r-1) = 10
10(r-1) = a(r^20 - 1)
(r-1) = (a/10)(r^20 - 1) - EQN 1

S30 = 91
a(r^30 - 1)/(r-1) = 91
Sub. EQN 1 into the above EQN:
a(r^30 - 1) * 10/a(r^20 - 1) = 91

THe solution just becomes more and more complicated... =/ is there something wrong? How do I simplify?
$\displaystyle \frac{a(r^{20} - 1)}{r - 1} = 10$ and $\displaystyle \frac{a(r^{30} - 1)}{r - 1} = 91$ and divide the second equation by the first. This will eliminate the $\displaystyle a$ and enable you to solve for $\displaystyle r$.