Question

if k+1,3k and 4k+2 be any three consecutive terms of an AP,find the value of k

Answer 1

A=K+1
D=3K-K-1
D=2K-1
AN=4K+2
AN=A+2D
4K+2=K+1+2(2K-1)
3K+2-1=4K-2
3K+1=4K-2
-K=-3
K=3

Answer 2

then a1=k+1
        a2=3k
        a3=4k+2
so a2-a1=a3-a2⇒d
  3k-k-1=4k+2-3k
     2k-1=k+2
          k=3

:)hope this ans would help u...