Question

determine k so that 4k+8,2k 2 +3k+6,3k 2 +4k+4 are three consecutive terms of an A.P.

Answer 1

we know according to AP
common difference is constant
now
2k^2+3k+6-4k-8=3k^2+4k+4-2k^2-3k-6
2k^2-k-2=k^2+k-2
k^2-2k=0
k (k-2)=0
k=0,2

Answer 2

Firstly please clear the ques. in second term,2k 2 +3k+6, here the second 2 is multiplied with 2k?