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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 )

Since the terms are in A.P

we have 2*(x^3 + 1) = x + x^4;; solving we get ;; (x-2)*(x+1) = 0

so we get values of x= 2 or -1 Hence possible sums of three terms will be

for x=2 will be { 2+9+16 ) = 27 for x= -1 will be {-1 + 0 + 1 } = 0

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

First and last term of a geometric progression are 3 and 96. If the sum of all these terms is 189, then find the number of terms in this progression.

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

An integer x is selected in such a way that

x + x^(1/2) + x^(1/4) = 276

Then what would be the sum of all digits of the number y where

y = 100000 x^(1/4) + 1000 x^(1/2) + x