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The sum upto terms of the series is
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a
a2+3nd2
b
a2+2nad+n(n−1)d2
c
a2+3nad+n(n−1)d2
d
(a+nd)2+n(n+1)d2
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Correct option is D
We can write the sum upto (2n+1) terms as[a+(a+d)](−d)+[(a+2d)+(a+3d)](−d)+…(a+(2n−2)d)+(a+(2n−1)d](−d)+(a+2nd)2 =(−d)[a+(a+d)+(a+2d)+… +a+(2n−1)d]+(a+2nd)2= (−d)2n2{a+a+(2n−1)d}+(a+2nd)2=−2nad−n(2n−1)d2+a2+4n(ad)+4n2d2=a2+2nad+n(2n+1)d2=(a+nd)2+n(n+1)d2Similar Questions
If the sum of first terms of an is , then the sum of squares of these terms is
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