If in an AP the sum of first x terms is Sx and Sx/Skx is independent of x ,then To proove d=2a

Answers (1)

Let "a" be the first term and "d" be the common difference.

Now SX/SKX=X/2 (2a + (X-1)d)/KX/2(2a + (KX - 1)d)

                      =(2a + Xd- d)/(2aK + K²Xd - Kd) 

                      =(2a + Xd- d)/[(2a-d)K + K²Xd]

                      =(2a + Xd- d)/k[(2a-d) + KXd]

Now since SX/SKX is independent of X then,

(2a - d )=0

or

d=2A

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