Question

What is the sum of the first five terms of a geometric series with a1=6 and r=1/3

Answer 1

The sum of n terms of a geometric series is found using the following formula:
$S _{n} = \frac{a(1-r^{n})}{1-r}$
where a is the first term and r is the common ratio.
In your question, a = 6 and r = 1/3, so we have:
$S_{5} =\frac{6(1-(\frac{1}{3})^5)}{1-\frac{1}{3}}$
which simplifies to:$S_{5}=\frac{6(1-\frac{1}{243})}{\frac{2}{3}}$