What is the least value of $n$ such that $(1 + 3 + 3^{2} + . . . . . + 3^{n})$ exceeds $2000$ ?

7

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5

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8

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6

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Solution

The correct option is A $7$
We know $S_{n} = \frac{a (r^{n} - 1)}{r - 1}$
Sum of the given G.P. is
$S_{n} = \frac{1 (3^{n + 1} - 1)}{3 - 1}$
Now according to question, we have
$\frac{1 (3^{n + 1} - 1)}{3 - 1} > 2000$
$\Rightarrow 3^{n + 1} - 1 > 4000$
$\Rightarrow 3^{n + 1} > 4001$
$\Rightarrow 3^{7} < 4001 < 3^{8}$
Therefore, $n = 7$

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