Question

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. Answer:

Answer 1

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

$\{x,xr,xr^2,xr^3\}$

We want to have

$\begin{cases}xr^2-x=12\\xr^3-xr=36\end{cases}$

In the second equation, we have

$xr^3-xr=xr(r^2-1)=36$

and in the first, we have

$xr^2-x=x(r^2-1)=12$

Substituting this into the second equation, we find

$xr(r^2-1)=12r=36\implies r=3$

So now we have

$\begin{cases}9x-x=12\\27x-3x=36\end{cases}\implies x=\dfrac32$

Then the four numbers are

$\left\{\dfrac32,\dfrac92,\dfrac{27}2,\dfrac{81}2\right\}$