SOLUTION: For which values of k are 2k, 5k+2, and 20k-4 consecutive terms of a geometric sequence?

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Question 934997: For which values of k are 2k, 5k+2, and 20k-4 consecutive terms of a geometric sequence?
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
%285k%2B2%29%2F2k+=+%2820k-4%29%2F%285k%2B2%29
2k(20k-4) = (5k+2)(5k+2) = 25K^2 + 20k + 4
40k^2 - 8k = 25K^2 + 20k + 4
15k^2 - 28k - 4 = 0
x+=+%2828+%2B-+sqrt%28+1024+%29%29%2F%2830%29+
x+=+%2828+%2B-+32%29%2F%2830%29+
k = 2, -2/15
and...checking
2k, 5k+2, and 20k-4
4, 12, 36, ...
-4/15, 4/3, -20/3,...