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by GmatTakerNo.1 » Mon May 03, 2010 7:38 am
If the first, third and thirteenth terms of an arithmetic progression are in geometric progression, and the sum of the fourth and seventh terms of this arithmetic progression is 40, find the first term of the sequence?

a) 2
b) 3
c) 4
d) 5
e) 6

The answer is A.

Can you please explain the different steps as well to solve the question.
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by liferocks » Mon May 03, 2010 7:45 am
Let a be the first term and d be the common diff

then a(a+12d)=(a+2d)^2
or a^2+12ad=a^2+4ad+4d^2
or 2a=d

also a+3d+a+6d=40
or 2a+9d=40
or d+9d=40
or d=4
hence a=d/2=2

Ans A
"If you don't know where you are going, any road will get you there."
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