Find four numbers that form a geometric progression such that the third term is greater than the first by 12 and the fourth is greater than the second by 36.

If is the first number in the progression, and is the common ratio between consecutive terms, then the first four terms in the progression are

We want to have

In the second equation, we have

and in the first, we have

Substituting this into the second equation, we find

So now we have

Then the four numbers are

$8

step-by-step explanation:

it is pretty much backwards.

you take the 4x2 multiply that, and that is = $8