Determine k, so that k2 + 4k + 8,


Determine k, so that k2 + 4k + 8, 2k2 + 3k + 6 and 3k2 + 4k + 4 are three consecutive terms of an AP.


Since, $k^{2}+4 k+8,2 k^{2}+3 k+6$ and $3 k^{2}+4 k+4$ are consecutive terms of an AP

$2 k^{2}+3 k+6-\left(k^{2}+4 k+8\right)=3 k^{2}+4 k+4-\left(2 k^{2}+3 k+6\right)=$ Common difference

$\Rightarrow \quad 2 k^{2}+3 k+6-k^{2}-4 k-8=3 k^{2}+4 k+4-2 k^{2}-3 k-6$

$\Rightarrow \quad k^{2}-k-2=k^{2}+k-2$

$\Rightarrow \quad-k=k \Rightarrow 2 k=0 \Rightarrow k=0$


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