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.

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.

Thanks for the reply!

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?

Re: Geometric Progression

Quote:

Originally Posted by

**Blizzardy** Thanks for the reply!

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?

You have

My advice is to multiply the numerator and denominator of the right hand side of this equation by r^10. Then factorise the denominator, then look for common factors to cancel, simplify etc.

I end up with the simple equation