What is the sum of the geometric series in which a1 = −5, r = 3, and an = −405?

1 answer:

4 0

$a_1=-5;\ r=3;\ a_n=-405\\\\a_n=a_1r^{n-1}\\\\subtitute\\\\-5\cdot3^{n-1}=-405\ \ \ |divide\ both\ sides\ by\ (-5)\\\\3^{n-1}=81\\\\3^{n-1}=3^4\iff n-1=4\to n=5\\\\S_n=\dfrac{a_1(1-r^n)}{1-r}\\\\S_5=\dfrac{-5(1-3^5)}{1-3}=\dfrac{-5(1-243)}{-2}=\dfrac{5\cdot(-242)}{2}=5\cdot(-121)=-605\\\\\\Answer:\boxed{S_5=-605}$

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