Answer
arithmetic,
$1650$
Work Step by Step
An arithmetic sequence has a common difference
$a_{n}=a+(n-1)d, $with nth partial sum:
1. $ S_{n}=\displaystyle \frac{n}{2}[2a+(n-1)d]\qquad$ or 2. $S_{n}=n(\displaystyle \frac{a+a_{n}}{2})$
A geometric sequence has a common ratio
$a_{n}=ar^{n-1}$, with nth partial sum:
$S_{n}=a\displaystyle \cdot\frac{1-r^{n}}{1-r}$
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Testing for common difference:
$\displaystyle \frac{2}{3}-\frac{1}{3}=\frac{1}{3},\displaystyle \qquad 1-\frac{2}{3}=\frac{1}{3},\quad\frac{4}{3}-1=\frac{1}{3}$
There is a common difference, the sequence is arithmetic,
$a=\displaystyle \frac{1}{3}$ and $d=\displaystyle \frac{1}{3}$
$a_{n}=\displaystyle \frac{1}{3}+\frac{1}{3}(n-1)$
The last term is $33$, from which we find n
(the number of terms in the sum)
$\displaystyle \frac{1}{3}+\frac{1}{3}(n-1)=33\qquad/\times 3$
$1+n-1=99$
$n=99$
So,
$S_{99}=n(\displaystyle \frac{a+a_{n}}{2})=\frac{99}{2}(\frac{1}{3}+\frac{99}{3})$
$=\displaystyle \frac{99}{2}\cdot\frac{100}{3}=1650$
