Find $3 + 12 + 48 + . . .$ up to $5$ terms.

1023

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2023

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3023

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4023

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Solution

The correct option is C $1023$
Given series is $3 + 12 + 48 + . . . .$ upto $5$ terms
Here $a = 3, r = 4, n = 5$
$S_{n} = \frac{a_{1} (r^{n} - 2)}{r - 1}$
$S_{5} = \frac{3 ((4)^{5} - 1)}{4 - 1}$
$= \frac{3 (1023)}{3}$
$= 1023$

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