$\bf \qquad \qquad \textit{sum of a finite geometric sequence}
\\\\
S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}
\end{cases}
\\\\\\
\sum\limits_{n=1}^{8}~-2(3)^{n-1}~~
\begin{cases}
n=8\\
a_1=-2\\
r=3
\end{cases}\implies S_8=-2\left( \cfrac{1-3^8}{1-3} \right)
\\\\\\
S_8=\underline{-2}\left( \cfrac{1-6561}{\underline{-2}} \right)\implies s_8=1-6561\implies S_8=-6560$

8 0

4 years ago

Solve the equation. Then check your solution.

Arte-miy333 [17]

8x+5= 69

Move +5 to the other. Sign changes from +5 to -5.

8x+5-5= 69-5

8x= 64

Divide by 8

8x/8= 64/8

x= 8

Answer : x= 8

4 0

3 years ago