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If a is the first term, d the common difference and Sk the sum to k terms of an A.P., then for to be independent of x
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a
a=2d
b
a=d
c
2a=d
d
None of these
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Correct option is C
We have, SkxSx=kx2[2a+(kx−1)d]x2[2a+(x−1)d]=k[(2a−d)+kxd](2a−d)+xdFor SkxSxto be independent of x, 2a – d = 0 or 2a = d.Similar Questions
If the sum of first terms of an is , then the sum of squares of these terms is
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