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

Given: $k^2+4k+8, 2k^2+3k+6,\ 3k^2+4k+4$ are three consecutive terms of an A.P.

To do: 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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Updated on: 10-Oct-2022

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