# Geometric Progression

• Jun 22nd 2011, 02:52 AM
Blizzardy
Geometric Progression
Hi guys, need help with this question please:

The first term of a geometric series is 8. The sum of its first 10 terms is 1/8 of the sum of the reciprocal of these terms. Find the common ratio.
• Jun 22nd 2011, 02:58 AM
mr fantastic
Re: Geometric Progression
Quote:

Originally Posted by Blizzardy
Hi guys, need help with this question please:

The first term of a geometric series is 8. The sum of its first 10 terms is 1/8 of the sum of the reciprocal of these terms. Find the common ratio.

x = geometric sum where a = 8, n = 10 and common ratio = r. Use the usual formula to get x in terms of r.

y = geometric sum where a = 1/8, n = 10 and common ratio = 1/r. Use the usual formula to get y in terms of r.

Substitute into x = (1/8)y and solve for r.
• Jun 22nd 2011, 07:29 PM
Blizzardy
Re: Geometric Progression
Quote:

Originally Posted by mr fantastic
x = geometric sum where a = 8, n = 10 and common ratio = r. Use the usual formula to get x in terms of r.

y = geometric sum where a = 1/8, n = 10 and common ratio = 1/r. Use the usual formula to get y in terms of r.

Substitute into x = (1/8)y and solve for r.

I substituted the values and ended up with this equation:
8(1-r^10) / (1-r) = (1/64)(1-r^-10) / (1-r^-1)
8/(1/64) = [(1-r^-10) / (1-r^-1)] / [(1-r^10) / (1-r)]

But the solution just becomes more complicated, how do I continue?
• Jun 22nd 2011, 07:59 PM
mr fantastic
Re: Geometric Progression
Quote:

Originally Posted by Blizzardy
You have $\displaystyle \frac{8(1 - r^{10})}{1 - r} = \frac{1}{64} \frac{1 - \frac{1}{r^{10}}}{1 - \frac{1}{r}}$
I end up with the simple equation $\displaystyle r^9 = - \frac{1}{8^3}$