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?

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Answer 1 · 2015-11-25T20:00:48

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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.