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

Here the given AP is

$k\textsuperscript{2} + 4k + 8, 2k\textsuperscript{2} + 3k + 6, 3k\textsuperscript{2} + 4k + 4\\$

$a\textsubscript{1}= k\textsuperscript{2} + 4k + 8, a\textsubscript{2}=2k\textsuperscript{2}+3k+6, a\textsubscript{3}=3k\textsuperscript{2} + 4k + 4\\$

$\\a\textsubscript{2}- a\textsubscript{1} = 2k\textsuperscript{2} + 3k + 6 - (k\textsuperscript{2} + 4k+ 8)\\ = 2k\textsuperscript{2} + 3k + 6 - k\textsuperscript{2} - 4k - 8\\ = k\textsuperscript{2} - k - 2\\ a\textsubscript{3} - a\textsubscript{2} = 3k\textsuperscript{2} + 4k + 4 - (2k\textsuperscript{2}+3k + 6) \\ = 3k\textsuperscript{2} + 4k + 4 - 2k\textsuperscript{2} - 3k - 6\\ = k\textsuperscript{2} + k - 2\\$

Hence, it is given that these terms are an AP

So the common difference of the A.P

$\\a\textsubscript{2} - a\textsubscript{1} = a\textsubscript{3} - a\textsubscript{2}\\ k\textsuperscript{2} - k - 2 = k\textsuperscript{2} + k - 2\\ k\textsuperscript{2} - k - 2 - k\textsuperscript{2} - k + 2 = 0\\ -2k = 0\\ k = 0\\$