Question

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

Answer 1

Let b=na. Since 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=n^2 a

Given that their arithmetic mean is b+2 (=na+2), it means:
(a + na + n^2 a)/3 = na+2

Simplifying,
n^2 -2n + (1 - 6/a) = 0

Finding the roots of this quadratic equation in n,
n =1 ± 0.5√(24/a) [using quadratic formula]
Since n is an integer, 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,
a²+ a -14/a+1
=6^2+6-14/7
=4