The side length of the square whose area is equal to the $7^{t h}$ term of the sequence $2, \sqrt{8}, 4, \dots,$ is

3

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4

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5

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6

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Solution

The correct option is B $4$
Given sequence $2, \sqrt{8}, 4, \dots$ is a G.P., whose common ratio is $r = \sqrt{2}$
$7^{t h}$ term is
$T_{7} = a r^{6} = 2 (\sqrt{2})^{6} = 16$

Therefore, the side length of the square is $4$ .

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