How do you find $S_{n}$ $S_n$ for the geometric series $a_{1} = 1$ $a_1=1$ , $a_{6} = - 243$ $a_6=-243$ , r=-3?

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Answer 1 · 2017-09-02T12:59:51

$- 182$ $-182$

firstly we need to write out the GP

$S_{n} = 1 + (- 3) + (9) + (- 27) + (81) + (- 243)$ $S_n=1+(-3)+(9)+(-27)+(81)+(-243)$

multiply the series by the common ratio $r = - 3$ $r=-3$

$- 3 S_{n} = (- 3) + (9) + (- 27) + (81) + (- 243) + (729)$ $-3S_n=" "color(red)((-3)+(9)+(-27)+(81)+(-243))+(729)$

$S_{n} = 1 + (- 3) + (9) + (- 27) + (81) + (- 243)$ $S_n" "=1+color(red)((-3)+(9)+(-27)+(81)+(-243))$

Subtract the two. It will be noticed that the terms in red will cancel.

$- 4 S_{n} = 729 - 1$ $-4S_n=729-1$

$S_{N} = \frac{728}{- 4} = - 182$ $S_N=728/-4=-182$

How do you find SnS_n for the geometric series a1=1a_1=1, a6=−243a_6=-243, r=-3?