Answer
The sequence is geometric.
$r=\dfrac{2}{3}$
$S_{50} \approx 2$
Work Step by Step
$\bf\text{RECALL:}$
$\bf\text{(1) Arithmetic Sequence }$
A sequence is arithmetic if there exists a common difference $d$ among consecutive terms.
$d=a_n-a_{n-1}$
The sum of the first $n$ terms of an arithmetic sequence is given by the formulas:
$S_n=\frac{n}{2}(a_1 +a_n)$
or
$S_n=\frac{n}{2}\left(2 a_1 + (n-1)d\right)$
$\bf\text{(2) Geometric Sequence }$
A sequence is geometric if there exists a common ratio $r$ among consecutive terms.
$r=\dfrac{a_n}{a_{n-1}}$
The sum of the first $n$ terms of a geometric sequence is given by the formula:
$S_{n}=a_1 \cdot \dfrac{1-r^n}{1-r}$
In the formulas listed above,
$d$ = common difference
$r$ = common ratio
$a_1$ = first term
$a_n$ = nth term
$n$ = number of terms
$\bf\text{List the first few terms of the sequence.}$
$\bf\text{Identify the sequence as arithmetic or geometric.}$
Substitute $1, 2, 3$ for $n$ to list the first three terms:
$a_1 =(\frac{2}{3})^1=\frac{2}{3}$
$a_2 = (\frac{2}{3})^2=\frac{4}{9}$
$a_3 = (\frac{2}{3})^3=\frac{8}{27}$
There is no common difference, so the sequence is not arithmetic.
Solve for the ratio of pairs of consecutive terms to obtain:
$\dfrac{a_2}{a_1} = \dfrac{\frac{4}{9}}{\frac{2}{3}}=\dfrac{4}{9} \cdot \dfrac{3}{2} = \dfrac{12}{18} = \dfrac{2}{3}$
$\require{cancel}\dfrac{a_3}{a_2} = \dfrac{\frac{8}{27}}{\frac{4}{9}}=\dfrac{8}{27} \cdot \dfrac{9}{4} = \dfrac{\cancel{8}2}{\cancel{27}3} \cdot \dfrac{\cancel{9}}{\cancel{4}}=\dfrac{2}{3}$
The sequence has a common ratio of $\dfrac{2}{3}$.
Thus, the sequence is geometric with $d=\frac{2}{3}$.
$\bf\text{Find the sum of the first 50 terms}:$
With $a_1=\dfrac{2}{3}$ and $r=\frac{2}{3}$, solve for the sum of the first 50 terms using the formula in (2) above to obtain:
$S_n = a_1 \cdot \dfrac{1-r^n}{1-r}
\\S_{50} = \dfrac{2}{3} \cdot \left(\dfrac{1-\cdot(\frac{2}{3})^{50}}{1-\frac{2}{3}}\right)
\\S_{50} \approx 2$
