# How to substitute to solve this Arithmetic progression problem?

• April 11th 2009, 10:01 PM
azuki
How to substitute to solve this Arithmetic progression problem?
The sum of the first 100 terms of an arithmetic progression is 5000, the first, second and fifth of this progression are three consectuive terms of a geometric progression. Find the value of the first term, and the non-zero common difference of the arithmetic progression.

I only manage to get
(100/2) (2a + 99d) = 5000----(1)
(a+d)^2 = a(a + 4d) ----(2)

I try to substitute, but I just can't get the answer. Am I moving in the wrong direction?
• April 11th 2009, 11:58 PM
red_dog
From (2): $a^2+2ad+d^2=a^2+4ad\Rightarrow 2ad=d^2\Rightarrow d=2a$

Now substitute 2a with d in (1).

Spoiler:
Answer: $d=1, \ a=\frac{1}{2}$