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=\frac32`

Then the four numbers are

`\left\{\frac32,\frac92,\frac{27}2,\frac{81}2\right\}`