Question

Find k so that the numbers 2k+1,3k+4,and 7k+6 form a geometric sequnce.

Answer 1

Let a, b and c be in a geometric sequence, then ac = b^2
Hence, (2k + 1)(7k + 6) = (3k + 4)^2
14k^2 + 19k + 6 = 9k^2 + 24k + 16
5k^2 - 5k - 10 = 0
5k^2 + 5k - 10k - 10 = 0
5k(k + 1) - 10(k + 1) = 0
(5k - 10)(k + 1) = 0
5k - 10 = 0 or k + 1 = 0
5k = 10 or k = -1
k = 2 or k = -1

The geometric sequence formed is
2(2) + 1, 3(2) + 4, and 7(2) + 6
5, 10, and 20
OR
2(-1) + 1, 3(-1) + 4, and 7(-1) + 6
-1, 1, and -1