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
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
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
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 )