## College Algebra (10th Edition)

The sequence is geometric. $r=\dfrac{2}{3}$ $S_{50} \approx 2$
$\bf\text{RECALL:}$ $\bf\text{(1) Arithmetic Sequence }$ 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: $S_n = a_1 \cdot \dfrac{1-r^n}{1-r} \\S_{50} = \dfrac{2}{3} \cdot \left(\dfrac{1-\cdot(\frac{2}{3})^{50}}{1-\frac{2}{3}}\right) \\S_{50} \approx 2$