Question

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?

Answer 1

See explanation...

Explanation:

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

`2x, 3x+4, 8x`

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

`3x+4 = +-sqrt(2x * 8x) = +-sqrt(16 x^2) = +-4x`

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

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

If `3x+4 = -4x` then `x = -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:

`2x = -8/7`

`3x+4 = -12/7+4 = 16/7`

`8x = -32/7`

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