Answer
The sequence is arithmetic.
$d=-\dfrac{2}{3}$
$S_{50} = -700$
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:
$a_1 =3-\frac{2}{3}(1)=3-\frac{2}{3} = \frac{9}{3} - \frac{2}{3} =\frac{7}{3}$
$a_2 = 3-\frac{2}{3}(2)=3-\frac{4}{3} = \frac{9}{3} - \frac{4}{3} =\frac{5}{3}$
$a_3 = 3-\frac{2}{3}(3)=3-\frac{6}{3} = \frac{9}{3} - \frac{6}{3} =\frac{3}{3}=1$
Notice that the values decrease by $\frac{2}{3}$.
Thus, the sequence is arithmetic with $d=-\frac{2}{3}$.
$\bf\text{Find the sum of the first 50 terms}:$
With $a_1=\dfrac{7}{3}$ and $d=-\frac{2}{3}$, solve for the sum of the first 50 terms using the formula in (1) above to obtain:
$S_n = \dfrac{n}{2}\left(2a_1+(n-1)d)\right)
\\S_{50} = \dfrac{50}{2}\left(2\cdot(\frac{7}{3}) + (-\frac{2}{3})(50-1)\right)
\\S_{50} = 25(\frac{14}{3}+(-\frac{2}{3}) \cdot 49)
\\S_{50} = 25(\frac{14}{3} -\frac{98}{3})
\\S_{50}=25(-\frac{84}{3})
\\S_{50}=25(-28)
\\S_{50}=-700$
