Question

If abc are positive integers forming an increasing gp and ba is perfect cube

Answer 1

Answer:

Step-by-step explanation:

Let a, b, c be positive integers such as b/a 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 a^2+a-14/a-1 is?

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,

a

2

+ a -14/a+1

=6^2+6-14/7

=4

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