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Let `S_n` denote the sum of n terms of an AP whose first term is a. If common difference d is given by `d=Sn-kS_(n-1)+S_(n-2)` , then k is : |
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Answer» `d = a_n - a_(n-1)` `a_n = s_n - s_(n-1)` `= n/2 [2a + (n-1)d] - (n-1)/2[2a + (n-2)d]` `= (2an + n(n-1)d - (n-1)[2a+ (n-1)d])/2` `= (2a+ (n-1)d-d +nd)/2` `= a + nd-d` `= a + (n-1)d` `k=2` so, option B is correct answer |
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