Question

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

Answer 1

2(3k+3) = (5k-3)+(2k+1)
⇒6k +6 = 7k -2
⇒k =8
acc to question it is 5k -1
so,
2(3k+3) = (5k-1)+(2k+1)
⇒6k +6 = 7k
⇒k =6