Answer
The sequence is geometric with $d=\dfrac{2}{3}$.
and
$S_{50} \approx 2$
Work Step by Step
We will check whether the sequence is arithmetic or geometric.
(1) A sequence is arithmetic if there exists a common difference $d$ among consecutive terms such that: $d=a_n−a_{n−1}$
The sum of the first n terms of an arithmetic sequence can be computed as:
$S_n=\dfrac{n}{2}[2a_1+(n−1)d] (1)$
(2) A sequence is geometric if there exists a common ratio $r$ among consecutive terms such that: $r=\dfrac{a_n}{a_{n−1}}$
The sum of the first n terms of a geometric sequence can be computed as:
$S_n= \dfrac{a_1(1-r^n)}{1-r} (2)$
where, $a_1 =\ First \ term$, $a_n$ = $n$th term, and $n =\ Number \ of \ Terms$
Substitute $n=1,2,3$ to list the first three terms:
$a_1 =(\dfrac{2}{3})^1=\dfrac{2}{3} \\ a_2 = (\dfrac{2}{3})^2=\dfrac{4}{9} \\ a_3 = (\dfrac{2}{3})^3=\dfrac{8}{27}$
We see that there is no common difference, so the sequence is not arithmetic.
Solve for the ratio of pairs of consecutive terms to obtain:
$\dfrac{a_2}{a_1} = \dfrac{\frac{4}{9}}{\frac{2}{3}} = \dfrac{2}{3}$
and $\dfrac{a_3}{a_2} = \dfrac{\frac{8}{27}}{\frac{4}{9}}=\dfrac{2}{3}$
We see that the sequence has a common ratio of $r=\dfrac{2}{3}$. So, the sequence is geometric with $d=\dfrac{2}{3}$.
In order to find the sum of the first 50 terms, we will substitute $a_1=\dfrac{2}{3}$ and $r=\dfrac{2}{3}$ into the formula in (2) to obtain:
$S_{50} = \dfrac{2}{3} \cdot \left(\dfrac{1-\cdot(\frac{2}{3})^{50}}{1-\frac{2}{3}}\right) \approx 2$
Therefore, the sum of first 50 terms is equal to: $S_{50} \approx 2$
