# Three integers are in the ratio 2:3:8. If 4 is added to the middle number, the resulting number is the second term of a geometric progression of which the other two integers are the first and third terms. How do you find the three integers?

Nov 25, 2015

See explanation...

#### Explanation:

If the first term is $2 x$ then the sequence formed by adding $4$ to the middle number is:

$2 x , 3 x + 4 , 8 x$

The middle of three terms of a geometric sequence is equal to the geometric mean of the preceding and following terms, so:

$3 x + 4 = \pm \sqrt{2 x \cdot 8 x} = \pm \sqrt{16 {x}^{2}} = \pm 4 x$

If $3 x + 4 = 4 x$ then $x = 4$ and our original integers are $8$, $12$, and $32$.

The geometric sequence is $8$, $16$, $32$ with common ratio $2$.

If $3 x + 4 = - 4 x$ then $x = - \frac{4}{7}$, but this is not an integer and does not give rise to integers when multiplied by $2$, $3$ and $8$.

Out of curiosity let's look at this alternative non-integer solution:

$2 x = - \frac{8}{7}$

$3 x + 4 = - \frac{12}{7} + 4 = \frac{16}{7}$

$8 x = - \frac{32}{7}$

So the common ratio of this geometric sequence is $- 2$ as expected.