Question

fin a1 in geometric series.
s12=3071.25 and r=2
a.0.5
b.0.75
c.2.5
d.3

Answer 1

\[S _{n}=\frac{ a(r^n-1) }{ r-1 }\]
\[S _{12}=\frac{ a(2^{12}-1) }{ 2-1 }\]
\[3071.25=\frac{ a(2^{12}-1) }{ 2-1 }\]
Solve for a by rearranging the expression.