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

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Solution

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

Therefore, the difference between any two consecutive numbers is same.

Hence, $(2 k^{2} + 3 k + 6) - (k^{2} + 4 k + 8) = (3 k^{2} + 4 k + 4) - (2 k^{2} + 3 k + 6)$

$⟹ k^{2} - k - 2 = k^{2} + k - 2$

$⟹ k + k = 0$

$⟹ 2 k = 0$

$⟹ k = 0$

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