Question

for what value of k will the consecutive terms 2k+1,3k+3,5k-3 form an AP?

Answer 1

For being it in AP, there must be common difference.
2(3k+3) = (5k-3)+(2k+1)
⇒6k +6 = 7k -2
⇒k =8

according to question it is 5k -1
so,
2(3k+3) = (5k-1)+(2k+1)
⇒6k +6 = 7k
⇒k =6

Answer 2

In ap middle term is defined by (a+c)/2=b Where a, b, c are consecutive term of ap So here (2k+1+5k-3)/2=3k+3 7k-2=6k+6 K=8