SOLUTION: The sum of the first 100 terms of an arithmetic progression is 10000; the first, second and fifth terms of this progression are three consecutive terms of a geometric progression.

Question 732799: The sum of the first 100 terms of an arithmetic progression is 10000; the first, second and fifth terms of this progression are three consecutive terms of a geometric progression. Find the first term, a, and the non-zero common difference, d. of the arithmetic progression
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
S 100 = 100/2[2a+99d]
S100 =50[2a+99d]
a, a+d, a+4d
(a+d)^2=a(a+4d)
a^2+2ad+d^2=a^2+4ad
d^2=2ad
d=2a
S100 =50[2a+99d]
S100= 50*100d
S100=5000d
10,000=5000d
d=2
Therefore a=1
Check
s100 = 100/2[2+198]
s100= 50*200
s100 = 10,000

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