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The minimum number of terms of the series  1+ 3 + 9 + 27 + … so that the sum may exceed 1000, is

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a
7
b
5
c
3
d
None of these

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detailed solution

Correct option is A

Let, the sum of n terms exceeds 1000.Then 11−3n1−3>1000; or 3n−12>1000 or 3n>2001; but 36=729 and 37=2187; ∴n>6;  but n is a positive integer∴n=7,8,9,…∴ The minimum number of terms = 7.


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