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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)
k-(1/2k) = (2k^2 - 1)/2k
Answer is 3/2
If x, (x^3+1) and x^4 are in arithmetic progression,then what will be possible sum of 3 terms of arithmetic progression?
( given that x is real number )
x, y, z and k are four non zero positive integers satisfying 1/x + y/2 = z/3 + 4/k, minimum integral value of k for integral value of x, y and z will be
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