Determine k so that $k^{2} + 4 k + 8 = 0, 2 k^{2} + 3 k + 6 = 0, 3 k^{2} + 4 k + 4 = 0$ are the three consecutive terms of an AP

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Solution

$k^{2} + 4 k + 8, 2 k^{2} + 3 k + 6, 3 k^{2} + 4 k + 4$ are the three consecutive terms of A.P
$\frac{k^{2} + 4 k + 8 + 3 k^{2} + 4 k + 4}{2} = 2 k^{2} + 3 k + 6 4 k^{2} + 8 k + 12 = 4 k^{2} + 6 k + 12 4 k - 3 k = 0 k = 0$

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A P .

Determine k so that k^2 + 4k + 8,2k^2 + 3k + 6,3k^2 + 4k + 4 are three consecutive terms of an A.P.

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