If k, k^2 and (2k+1) are in arithmetic progression then possible value of k-(1/2k) is?

Question

1) 1/2
2) 1
3) 3/2
4) 2
5) 5/2
6) 5/3
7) 5/4
8) 7/6
9) 6/13
10) 3/16

Answer 1

as its an arthmatic progressions

k^2 - k = (2k+1) - k^2
2k^2 - 3k = 1
2k^2 - 1 = 3k
divide by 2k on both side
(2k^2 - 1)/2k = 3/2--------------(1)
for
k-(1/2k) = (2k^2 - 1)/2k
from 1,
Answer is 3/2