Determine the value of k for which $k^{2} + 4 k + 8, 2 k^{2} + 3 k + 6$ and $3 K^{2} + 4 k + 4$ are in A.P.

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Solution

Let $k^{2} + 4 k + 8, 2 k^{2} + 3 k + 6$ and $3 k^{2} + 4 k + 4$ are in A.P.

$2 k^{2} + 3 k + 6 = \frac{k^{2} + 4 k + 8 + 3 k^{2} + 4 k + 4}{2} 4 k^{2} + 6 k + 12 = 4 k^{2} + 8 k + 12 6 k = 8 k 2 k = 0 k = 0$

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