Question

What is the sum of the geometric series in which a1 = 3, r = 2, and an = 192?

Answer 1

\[a _{n}=a _{1}r ^{n-1}\]
\[192=3*2^{n-1}\]
\[2^{n-1}=\frac{ 192 }{ 3 }=64=2^6\]
n-1=6
n=6+1=7
\[S _{n}=a _{1}\frac{ r^n-1 }{ r-1 }\]
find \[S _{7}\]

Answer 2

thats not the answer

Answer 3

\[S _{7}=3\frac{ 2^7-1 }{ 2-1 }=3*\frac{ 128-1 }{ 2-1 }=3*127=?\]