Question

find non zero value of k if k^2+4k+8,2k^2+3k+6,3k^2+4k+4 are 3 consecutive terms in a.p

Answer 1

$2k^2+3k+6=\dfrac{k^2+4k+8+3k^2+4k+4}{2}\\\\4k^2+6k+12=4k^2+8k+12\\2k=0\\k=0$

There is no non-zero value of $k$

Answer 2

There is no non zero value of k.

Step-by-step explanation:

The $k^{2} +4k+8,2k^{2} +3k+6 and 3k^{2} +4k+8$ are three consecutive terms in a.p.

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

⇒$2k^{2} +3k+6 - k^{2} -4k-8= 3k^{2} +4k+8 -2k^{2} -3k-6$

⇒$k^{2} -k -2= k^{2} +k -2$

⇒$k+k=-2+2$

⇒$2k=0$

⇒$k=0$, there is no non zero value of k.

Hence, there is no non zero value of k.