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 AP.
Given:
$k^2+4k+8, 2k^2+3k+6,\ 3k^2+4k+4$ are three consecutive terms of an A.P.
To do:
We have to find the value of $k$.
Solution:
As given $k^2+4k+8, 2k^2+3k+6,\ 3k^2+4k+4$ are three consecutive terms of an A.P.
Then, $( 2k^2+3k+6)-( k^2+4k+8)=( 3k^2+4k+4)-( 2k^2+3k+6)$
$\Rightarrow 2k^2+3k+6-k^2-4k-8=3k^2+4k+4-2k^2-3k-6$
$\Rightarrow k^2-k-2=k^2+k-2$
$\Rightarrow -k-2=k-2$
$\Rightarrow k+k=-2+2$
$\Rightarrow 2k=0$
$\Rightarrow k=0$
Thus, $k=0$
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