Question

For what value of k will the consecutive terms 2k+1 3k+3 5k-1 from an AP

Answer 1

a2-a1=a3-a2

3k+3-(2k+1)=5k-1-(3k+3)

3k+3-2k-1 =5k-1-3k-3

k+2=2k-4

transposing k to RHS and -4 to LHS

we get

2+4=2k-k

6=k