<PRE><B>Question 472701</B>
Let a-k, a, a+k be the three numbers which sum up to 24. It should appear evident that a = 8.


So our numbers are 8-k, 8, and 8+k. We want 7-k, 6, 8+k to be in a geometric progression. We know that the ratio between successive terms is constant, so


*[tex \LARGE \frac{6}{7-k} = \frac{8+k}{6}]


*[tex \LARGE 36 = (7-k)(8+k) = -k^2 -k + 56]


*[tex \LARGE k^2 + k - 20 = 0]


k = 4 or k = -5. Hence the sequences are 4,8,12 and 13,8,3 (both work).</PRE>