Question

What is the sum of the first 11 terms of the geometric sequence: -4, 12, -36, 108,...

Answer 1

$a_1=-4,\ a_2=12,\ a_3=-36,\ a_4=108,\ ...\\\\a_2:a_1=12:(-4)=-3\\a_3:a_2=-36:12=-3\\a_4:a_3=108:(-36)=-3\\a_{n+1}:a_n=-3=const.\\\\\text{It's a geometric sequence}\\\\a_n=a_1r^{n-1}\to a_n=-4\cdot(-3)^{n-1}=-4\cdot(-3)^{-1}\cdot(-3)^n\\\\=-4\cdot\left(-\dfrac{1}{3}\right)\cdot(-3)^n=\dfrac{4}{3}(-3)^n\\\\\text{The formula of a Sum of the First n Terms of a Geometric Sequence:}$

$S_n=\dfrac{a_1(1-r^n)}{1-r}\\\\\text{We have:}\\a_1=-4,\ r=-3,\ n=11\\\\\text{Substitute:}\\\\S_{11}=\dfrac{-4[1-(-3)^{11}]}{1-(-3)}=\dfrac{-4(1+177147)}{4}=-177148\\\\Answer:\ -177,148$