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Q. Let a, b, c be positive integers such that b/a is an integer. If a, b, c are in geometric progression and the arithmetic mean of a, b, c is b+2, the value of (a²+a-14) /(a+1) is..?

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  • $\begingroup$ Welcome to Mathematics StackExchange! Nice question. Have you tried solving it? Did you encounter any difficulties? $\endgroup$
    – Matti P.
    Commented Oct 22, 2019 at 9:40
  • $\begingroup$ Let $a = x$, $b=xd$, and $c=xd^2$. So $d^2-2d+1=\frac6x\to(d-1)^2=\frac6x$ $\endgroup$
    – Rushabh Mehta
    Commented Oct 22, 2019 at 9:41
  • $\begingroup$ What did you try? Where did you get stuck? $\endgroup$
    – Allawonder
    Commented Oct 22, 2019 at 9:59

1 Answer 1

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Write $d=b/a$ so $a+b+c=3ad+6=a(1+d+d^2)$, i.e. $6=(1-d)^2a$. The only square dividing $6$ is $1$, so $a=6$. Hence $\frac{a^2+a-14}{a+1}=a-\frac{14}{a+1}=6-2=4$.

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