## College Algebra (10th Edition)

The sequence is arithmetic. $d=-\dfrac{2}{3}$ $S_{50} = -700$
$\bf\text{RECALL:}$ $\bf\text{(1) Arithmetic Sequence }$ 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: $S_n = \dfrac{n}{2}\left(2a_1+(n-1)d)\right) \\S_{50} = \dfrac{50}{2}\left(2\cdot(\frac{7}{3}) + (-\frac{2}{3})(50-1)\right) \\S_{50} = 25(\frac{14}{3}+(-\frac{2}{3}) \cdot 49) \\S_{50} = 25(\frac{14}{3} -\frac{98}{3}) \\S_{50}=25(-\frac{84}{3}) \\S_{50}=25(-28) \\S_{50}=-700$