Answer
Arithmetic;
$1650$
Work Step by Step
Let's note by $\{a_n\}$ the given sequence. The sum of its first $k$ terms is:
$$S_k=\dfrac{1}{3}+\dfrac{2}{3}+1+\dfrac{4}{3}\dots+33.$$
The terms of the sequence are:
$$\begin{align*}
a_1&=\dfrac{1}{3}\\
a_2&=\dfrac{2}{3}\\
a_3&=1\\
a_4&=\dfrac{4}{3}\\
&\dots\\
a_k&=33.
\end{align*}$$
We notice that the common difference of consecutive terms is:
$$\dfrac{2}{3}-\dfrac{1}{3}=1-\dfrac{2}{3}=\dfrac{4}{3}-1=\dfrac{1}{3},$$
therefore constant, so the sequence is arithmetic with the elements:
$$\begin{cases}
a_1=\dfrac{1}{3}\\
d=\dfrac{1}{3}.
\end{cases}$$
Before calculating $S_k$ we must find $k$ (the number of terms in the sum):
$$\begin{align*}
a_k&=a_1+(k-1)d\\
33&=\dfrac{1}{3}+(k-1)\left(\dfrac{1}{3}\right)\\
99&=1+k-1\\
k&=99.
\end{align*}$$
Calculate the partial sum $S_{99}$ using the formula:
$$S_n=\dfrac{n(a_1+a_n)}{2}.$$
For $n=99$ we have:
$$S_{99}=\dfrac{99\left(\dfrac{1}{3}+33\right)}{2}=1650.$$
