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Question

Let a,b,c be positive integers such that ba is an integer. If a,b,c are in geometric progression and the arithmetic mean of a,b,c is b+2, then the value of a2+a14a+1 is

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Solution

We know that,
a+b+c3=b+2a+c=2(b+3)
We know that a,b,c are in G.P.
Let the common ratio be rI
a+ar2=2(ar+3)a=61+r22ra=6(r1)2
As both a and r, will be integer so only possible solution will be,
r=2a=6
Now,
a2+a14a+1=36+6147=4

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