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a
310
b
300
c
320
d
330
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Correct option is A
nth term of the series is 20+(n−1)(−2/3) For the sum to be maximum nth term ≥0⇒20+(n−1)−23≥0Thus, the sum of 31 terms is maximum and is 31240+30×−23=310Similar Questions
If the sum of first terms of an is , then the sum of squares of these terms is
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