Question

Three positive integers a,b and c are such that they are in G.P. with b/a being an integer. The arithmetic mean of a,b and c is b + 2, then find the value of (a^2 + a - 14)/a+1. Answer is an integer between 0 and 9 (inclusive).

Answer 1

Dear student
Let b=na. Since $\frac{b}{a}$ is an integer and both b and a are positive, n is also a positive integer.

Since the numbers are in geometric progression, it follows that:
c=n² a

Given that their arithmetic mean is b+2 (=na+2), it means:
$\frac{(a+na+n^{2}a)}{3}=na+2$

Simplifying,
$n^2-2n+\left(1-\frac{6}{a}\right)=0$

Finding the roots of this quadratic equation in n,
n =1 ± 0.5$\sqrt{\frac{24}{a}}$ [using quadratic formula]
Since n is an integer, $\frac{24}{a}$ has to be a perfect square. Hence the only possible value of a is 6, which gives n=2

So the values of a, b, c are 6, 12, 24, respectively.
Now putting the values we get,
$\frac{a^2+a-14}{a+1}$
=6²+6-$\frac{14}{7}$
=4