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?
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.