Question

for what value of k will be consecutive terms 2k+1,3k+3and 5k-1 from an ap

Answer 1

let
a1 = 2k+1
a2= 3k+3
a3 = 5k-1
given a1,a2,a3 are in ap

therefore
a2- a1 = a3-a2
3k+3-(2k+1) = 5k-1- (3k+3)

3k+3-2k-1 =5k-1-3k-3
k+2 =2k-4

k-2k =-4-2
-k = -6
k= 6

Answer 2

Condition for AP
2b=a+c
2*(3k+3)=2k+1+5k-1
6k+6= 7k
k=6
The required AP=13,21,29