Question

for what value of k will the consecutive terms 2k+1 3k+3 and 5 k-1 form an ap

Answer 1

in series in AP means
common difference always constant .
e.g
(3k + 3 ) -( 2k +1 ) = ( 5k - 1) -( 3k + 3)

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

K + 2 = 2k -4

-k = -6

K = 6

Answer 2

As it is given that the terms are in AP,there will be a common difference, d

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=6