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

See 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.