Finding common ratio given two sums in geometric series

[odolwa99]

Homework Statement

Q.: The sum of the first five terms of a geometric series is 5 and the sum of the next five terms is 1215. Find the common ratio of this series.

Homework Equations

Sn = $\frac{a(r^n - 1)}{r - 1}$

The Attempt at a Solution

a + ar + ar^2 + ar^3 + ar^4 = 5
ar^5 + ar^6 + ar^7 + ar^8 + ar^9 = 1215

ar^5 + ar^6 + ar^7 + ar^8 + ar^9 = 1215
-(a + ar + ar^2 + ar^3 + ar^4) = 5
r^5 + r^5 + r^5 + r^5 + r^5 = 1210

5r^5 = 1210
r^5 = 242
r = $\sqrt[5]{242}$
r $\approx$ 3

Answer: From textbook: 3

Please note that $\sqrt[5]{243}$ is exactly 3. My answer is close but still off the mark. Can someone help me figure out how to fix this? Thank you.

 

[dynamicsolo]

What if you factor r⁵ from your equation for 1215 and then divide it by the other equation? (Your approach is not algebraically correct.)

 

[odolwa99]

Ok, here it is...

r^5(a + ar + ar^2 + ar^3 + ar^4) = 1215
a + ar + ar^2 + ar^3 ar^4 = 5

r^5 = $\frac{1215}{5}$

r^5 = 243
r = $\sqrt[5]{243}$
r = 3

That works out. Thank you very much.