How do you find $S_{n}$ $S_n$ for the geometric series $a_{2} = - 36$ $a_2=-36$ , a_5=972#, n=7?

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How do you find $S_{n}$ $S_n$ for the geometric series $a_{2} = - 36$ $a_2=-36$ , a_5=972#, n=7?

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Answer 1 · 2017-01-18T09:36:16

$S_{n} = 3^{(n + 1)} {(- 1)}^{n} - 3$ $S_n=3^((n+1))(-1)^n-3$

Explanation:

The $n^{t h}$ $n^(th)$ term of a Geometric Series and its sum $S_{n}$ $S_n$ up to n^(th) $t e r m$ $term$ a_n $, w h o s e f i r s t t e r m i s$ $, whose first term is$ a_1 $and c o m m o n r a t i o i s$ $and common ratio is$ r# is given by

$a_{n} = a_{1} \times r^{(n - 1)}$ $a_n=a_1xxr^((n-1))$ and $S_{n} = a_{1} \times \frac{r^{n} - 1}{r - 1}$ $S_n=a_1xx(r^n-1)/(r-1)$

as $a_{2} = a_{1} \times r = - 36$ $a_2=a_1xxr=-36$ and $a_{5} = a_{1} \times r^{4} = 972$ $a_5=a_1xxr^4=972$

Dividing latter by former, we get $r^{3} = - \frac{972}{36} = - 27$ $r^3=-972/36=-27$

and hence $r = \sqrt[3]{- 27} = - 3$ $r=root(3)(-27)=-3$

and $a_{1} = \frac{36}{- 3} = - 12$ $a_1=36/-3=-12$

Hence, $S_{n} = - 12 \times \frac{{(- 3)}^{n} - 1}{(- 3) - 1} = - 12 \times \frac{{(- 3)}^{n} - 1}{- 4}$ $S_n=-12xx((-3)^n-1)/((-3)-1)=-12xx((-3)^n-1)/(-4)$ i.e.

$S_{n} = 3 ({(- 3)}^{n} - 1) = 3 (3^{n} {(- 1)}^{n} - 1) = 3^{(n + 1)} {(- 1)}^{n} - 3$ $S_n=3((-3)^n-1)=3(3^n(-1)^n-1)=3^((n+1))(-1)^n-3$

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How do you find $S_{n}$ $S_n$ for the geometric series $a_{2} = - 36$ $a_2=-36$ , a_5=972#, n=7?

1 Answer

Explanation:

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How do you find SnS_n for the geometric series a2=−36a_2=-36, a_5=972#, n=7?

1 Answer

Explanation:

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How do you find $S_{n}$ $S_n$ for the geometric series $a_{2} = - 36$ $a_2=-36$ , a_5=972#, n=7?