If Arithmetic Mean of 1st k terms of Arithmetic Progression with 1st term a and common difference d is equal to the Geometric Mean of 1st k terms of Geometric Progression with 1st term a and common ratio r, then find the range of r for which a≈2d/(r-1).

The equation for the Arithmetic Mean with an Arithmetic Progression with a common difference d for k terms with first term equal to a is

AM = (1/k)* [(a + d*0) + ... + (a + d*(k-1))]

The equation for the Geometric Mean with a Geometric Progression with a common ratio r for k terms with first term equal to a is

GM = [a*(r⁰)*...* a*(r^k)]^(1/k)

Given that AM = GM ...

(1/k)*[(a+d*0)+...+(a+d*(k-1))] = [a*r⁰*...*a*r^(k-1)]^(1/k)

and, if I did my math correctly, ...

a = [d*0 + ... + d*(k-1)]/[k*(r⁰ * ... * r^(k-1))^(1/k) - 1]

... Now where do I go from here???

Where k = 2, (again, if I did my math correctly),

a = d/[2*r^(1/2) - 1]

...Now where do I go???

[equation corrected as pointed out by brianjn...I am guessing my slight dyslexia caused me to invert the fraction]

Edited on January 29, 2008, 10:38 am

Posted by Dej Mar on 2008-01-25 11:49:00