Which term of the GP 1, 2, 4, 8,.... is 512?

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Solution

The correct option is C
10

Given the GP: 1, 2, 4, 8, ......

$\Rightarrow a = 1, r = 2$
The $n^{th}$ term of a GP of first term $a$ and common ratio $r$ is given by
$a_{n} = a r^{n - 1} .$
Let $a_{n} = 512.$
Then, $1 \times 2^{n - 1} = 512.$
i.e., $2^{n - 1} = 2^{9}$
$\Rightarrow n - 1 = 9$
$\Rightarrow n = 10$

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