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

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Solution

Since in an AP the difference between the term and the preceding term is always same.
Therefore,
$(2 k^{2} + 3 k + 6) - (k^{2} + 4 k + 8) = (3 k^{2} + 4 k + 4) - (2 k^{2} + 3 k + 6)$
$\Rightarrow k^{2} - k - 2 = k^{2} + k - 2$
$\Rightarrow 2 k = 0$
$\Rightarrow k = 0$

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

k^{2} + 4 k + 8 = 0, 2 k^{2} + 3 k + 6 = 0, 3 k^{2} + 4 k + 4 = 0

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