3 rational numbers form a GP, if 8 is added to the middle number, then this new sequence forms an AP...

Question

3 rational numbers form a geometric progression in its current order.

1. If 8 is added to the middle number, then this new sequence forms an arithmetic progression.
2. If 64 is later added to the last number, then this yet another sequence forms a geometric progression again.
3. First number of these sequences is an integer.

Find the numbers?

Answer 1

Let a,b,c be the numbers in GP.
Then we have the geometric mean = (a+c)^(0.5) = b
Then the terms become a, b+8, c which is in AP
Then arithmetic mean = (a+c)/2 = b+8.
Then the terms become a, b+8, c+64
Then geometric mean = (a+(c+64))^(0.5) = b+8

Now we have the solution as

a, (a + ((64a)^0.5/2)), (a + ((64a)^0.5/2))^(2)/a
Using trial and error and some common sense (Lol) for a=4 we get
a=4, b=12, c=36 which is in GP with 3 as the common ratio
Then
a=4, b=20, c=36 which is in AP with 16 as the common difference.
Then
a=4, b= 20, c=100 which is again in GP with 5 as the common ratio.

So a=4, b=12 & c=36 is a solution for the question.