How do you find $S_{n}$ $S_n$ for the geometric series $a_{1} = 12$ $a_1=12$ , $a_{5} = 972$ $a_5=972$ , r=-3?

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How do you find $S_{n}$ $S_n$ for the geometric series $a_{1} = 12$ $a_1=12$ , $a_{5} = 972$ $a_5=972$ , r=-3?

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Answer 1 · 2017-08-30T12:02:33

$S_{n} = \frac{12 (1 - {(- 3)}^{n})}{4}$ $S_n = (12(1-(-3)^n))/4$

Explanation:

The sum to $n$ $n$ of a geometric series, denoted as $S_{n}$ $S_n$ , is given by the formula $S_{n} = \frac{a (1 - r^{n})}{1 - r}$ $S_n = (a(1-r^n))/(1-r$ where $a$ $a$ is the first term of the series and $r$ $r$ is the common ratio.

$S_{n} = \frac{12 (1 - {(- 3)}^{n})}{1 - - 3} = \frac{12 (1 - {(- 3)}^{n})}{4}$ $S_n = (12(1-(-3)^n))/(1--3 )= (12(1-(-3)^n))/4$

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How do you find $S_{n}$ $S_n$ for the geometric series $a_{1} = 12$ $a_1=12$ , $a_{5} = 972$ $a_5=972$ , r=-3?

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Explanation:

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How do you find SnS_n for the geometric series a1=12a_1=12, a5=972a_5=972, r=-3?

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Explanation:

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How do you find $S_{n}$ $S_n$ for the geometric series $a_{1} = 12$ $a_1=12$ , $a_{5} = 972$ $a_5=972$ , r=-3?