Question

Determine the value of k for which ksquare+4k+8,2ksquare+3k+6 and 3ksquare+4k+4 are in A.P

Answer 1

$Heya$‼

$A.P. k^2+4k+8, 2k^2+3k+6 \: and \: 3k^2+4k+4$

$First\:term, a = k^2+4k+8$

$Now,$
$Common \:difference, d = a2-a1 \:or\: a3-a2$

$a2-a1=a3-a2$

$(2k^2+3k+6 )-( k^2+4k+8)=( 3k^2+4k+4)-( 2k^2+3k+6)$

$2k^2+3k+6-k^2-4k-8= 3k^2+4k+4- 2k^2-3k-6$

$k^2-k-2=k^2+k-2$

$k^2-k^2=k+k+2-2$

$0=2k+0$

$2k=0$

$k=\frac{0}{2}$

$k=0$


Hence, the value of k is 0 .