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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___


Solution

i) If a, b, c are in GP, then they can be taken as a,ar,ar2 where r,(r0) is the common ratio.
ii) Arithmetic mean of x1,x2,xn=x1+x2++xnn
Let a, b, c be a,ar,ar2 where rϵN
Also, a+b+c3=b+2
a+ar+ar2=3(ar)+6
ar22ar+a=6
(r1)2=6a
Since, 6a must be perfect square and aϵN
So, a can be 6 only
r1=±1r=2
and a2+a14a+1=36+6147=4

Mathematics

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