$Find the least positive integer such that 1 + 7 + 7^{2} + 7^{3} + . . . . + 7^{n} > 3500$

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Solution

The correct option is D 6
$It is given that S_{n} > 3500 Given series is in GP \Rightarrow \frac{a (r^{n} - 1)}{r - 1} > 3500 \Rightarrow \frac{1 (7^{n} - 1)}{7 - 1} > 3500 \Rightarrow 7^{n} - 1 > 21000 \Rightarrow 7^{n} > 21001 \Rightarrow Least positive n = 6 (as 7^{6} = 117649 and 7^{5} = 16807)$

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