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

+1 vote
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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

posted Mar 4, 2015 by Ankur Athari

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1 Answer

+1 vote

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

answer Mar 14, 2015 by Jaspalsingh Parmar



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