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

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Solution

If $a, b, c$ are three consecutive terms of an A.P then $2 b = a + c$

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

$4 k^{2} + 6 k + 12 = 4 k^{2} + 8 k + 12$

$6 k = 8 k$

$∴ k = 0$

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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.

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

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