Question 859032: the sum of the first and the third terms of a geometric progression is 40. the sum of the second and the fourth terms is 120. Determine the first term and the common ration of the progression.
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! let the terms be a,ar,ar^2,ar^3
First condition
a+ar^2=40
a(1+r^2)=40
(1+r^2)=40/a
II condition
ar+ar^3=120
ar(1+r^2)=120
substitute (1+r^2)
ar(*(40/a) = 120
40r= 120
r=3
a(1+r^2)= 40
substitute r
a(1+9)=40
10a=40
a=4
CHECK
4, 12, 36,108,
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