Determine k so that k^{2}+4k+8, 2k^{2}+3k+6, | vedclass

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(D) Since the terms $k^{2}+4k+8, 2k^{2}+3k+6,$ and $3k^{2}+4k+4$ are in an $AP$,the difference between consecutive terms must be equal.Let the terms b

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(D) Since the terms $k^{2}+4k+8, 2k^{2}+3k+6,$ and $3k^{2}+4k+4$ are in an $AP$,the difference between consecutive terms must be equal.Let the terms be $a_1, a_2, a_3$.Then,$a_2 - a_1 = a_3 - a_2$.Substituting the values:$(2k^{2}+3k+6) - (k^{2}+4k+8) = (3k^{2}+4k+4) - (2k^{2}+3k+6)$Simplifying both sides:$k^{2} - k - 2 = k^{2} + k - 2$Subtracting $k^{2}$ and adding $2$ to both sides:$-k = k$$2k = 0$$k = 0$

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

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