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+a−14a+1 is___

Solution

## i) If a, b, c are in GP, then they can be taken as a,ar,ar2 where r,(r≠0) 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 ⇒ar2−2ar+a=6 ⇒(r−1)2=6a Since, 6a must be perfect square and aϵN So, a can be 6 only ⇒r−1=±1⇒r=2 and a2+a−14a+1=36+6−147=4Mathematics

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