Find $K$
If $K^{2} + 4 K + 8, 2 K^{2} + 3 K + 6$ and $3 K^{2} + 4 K + 4$ are any $3$ consecutive terms of $A . P$

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Solution

$K^{2} + 4 K + 8, 2 K^{2} + 3 K + 6, 3 K^{2} + 4 K + 4$
$K^{2} + 4 K + 8 = a$
$2 K^{2} + 3 K + 6 = b$
$3 K^{2} + 4 K + 4 = c$
$2 b = a + c$
$4 K^{2} + 6 K + 12 = 4 K^{2} + 8 K + 12$
$2 K = 0$
$K = 0$
$∴ K = 0$

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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.

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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