Practice Solutions

1. Determine the common ratio.

$r=\frac{a_2}{a_1}=\frac{–10}{5}=–2$

$r=–2$

2. Find $a_7, a_{10}$.

$$\begin{gathered} a_1=5, r=–2 \\ {a}_n=a_1·{\left(r\right)}^{n–1} \\ {a}_7=5{\left(–2\right)}^{7–1}=5{\left(–2\right)}^6=320 \\ {a}_{10}=5{\left(–2\right)}^{10–1}=5{\left(–2\right)}^9=–2560 \end{gathered}$$

$$\begin{gathered} a_7=320 \\ {a}_{10}=–2560 \end{gathered}$$

3. Find $a_1$.

$\begin{array}{rcl}{a}_n& =& \left(a_1\right)·r{ }^{n–1}\\ a_6& =& \left(a_1\right)·r{ }^{6–1}\\ 9& =& \left(a_1\right){\left(\frac{1}{3}\right)}^{6–1}\\ 9& =& \left(a_1\right){\left(\frac{1}{3}\right)}^5\\ \left(243\right)9& =& \left(a_1\right)\left(\frac{1}{243}\right)\left(243\right)\\ a_1& =& 2187\end{array}$

$a_1=2187$

4. Write the first three terms of the sequence.

$$\begin{gathered} a_1=2187, r=\frac{1}{3} \\ {a}_1=2187 \\ {a}_2=2187\left(\frac{1}{3}\right)=729 \\ {a}_3=729\left(\frac{1}{3}\right)=243 \end{gathered}$$

$\left\{2187, 729, 243\right\}$

5. Determine the amount of medication left in a person’s body after 3 hours.

$$\begin{gathered} a_1=100 \\ {a}_2=100\left(0.85\right)=85 \\ {a}_3=85\left(0.85\right)=72.25 \end{gathered}$$

After three hours there is 72.25 mg of medication left in a person’s body.

6. After approximately how long will a person have 50 milligrams of medication remaining in their body?

$$\begin{gathered} a_n=50 \\ {a}_1=100 \\ r=0.85 \\ {a}_n=a_1{\left(r\right)}^{n–1} \\ 50=100{\left(0.85\right)}^{n–1} \\ 0.5={\left(0.85\right)}^{n–1} \\ \log \left(0.5\right)=\left(n–1\right) \log \left(0.85\right) \\ \frac{\log \left(0.5\right)}{\log \left(0.85\right)}=n–1 \\ n=\frac{\log \left(0.5\right)}{\log \left(0.85\right)}+1 \\ n=5.27 \end{gathered}$$

After 5.27 hours, a person will have 50 mg of medication remaining in their body.

7. $a_1=0.8, r=2.1, S_{12}=\underline{ }$

$$\begin{gathered} S_{12}=\frac{a_1\left(1–r^{12}\right)}{1–r} \\ {S}_{12}=\frac{0.8\left(1–2.1^{12}\right)}{1–2.1}=5348.9654 \end{gathered}$$

$S_{12}=5348.97$

8. $\sum _{k=1}^5{\left(\frac{1}{2}\right)}^{k–2}$

$$\begin{gathered} a_1={\left(\frac{1}{2}\right)}^{1–2}={\left(\frac{1}{2}\right)}^{–1}=2 \\ r=\frac{1}{2} \\ n=5 \\ {S}_5=\frac{2\left(1–{\left({\displaystyle \frac{1}{2}}\right)}^5\right)}{1–{\displaystyle \frac{1}{2}}}=3.875 \end{gathered}$$

$S_5=3.88$

9. $S_8=1640, r=3, a_1=\underline{ }$

$\begin{array}{rcl}{S}_8& =& \frac{a_1\left(1–3^8\right)}{1–3}\\ \left(–2\right)1640& =& \frac{a_1\left(1–3^8\right)}{\cancel{–2}}\cancel{\left(–2\right)}\\ \frac{–3280}{\left(1–3^8\right)}& =& \frac{a_1\cancel{\left(1–3^8\right)}}{\cancel{\left(1–3^8\right)}}\end{array}$

$\begin{array}{rcl}{a}_1& =& 0.5\end{array}$

10. $\sum _{k=–1}^6{\left(\frac{5}{4}\right)}^k$

$$\begin{gathered} a_1={\left(\frac{5}{4}\right)}^{–1}=\frac{4}{5} \\ r=\frac{4}{5} \\ n=6–\left(–1\right)+1=8 \\ {S}_8=\frac{{\displaystyle \frac{4}{5}}\left(1–{\left({\displaystyle \frac{{\displaystyle 5}}{4}}\right)}^8\right)}{1–{\displaystyle \frac{{\displaystyle 5}}{4}}}=15.873 \end{gathered}$$

$S_8=15.87$

11. Calculate the total savings for a year.

$$\begin{gathered} a_1=100 \\ r=1+0.03=1.03 \\ n=12 \\ \begin{array}{c}{S}_{12}=\frac{100\left(1–1.{03}^{12}\right)}{1–1.03}\end{array}=1419.21 \end{gathered}$$

Kristin saved $1,419.21 after one year (12 months).

12. If Kristin has continued the pattern of saving and currently has $8,202.32 in her account, how long has she been saving?

$$\begin{gathered} \begin{array}{rcl}& & \begin{array}{c}{a}_1=100 \\ r=1.03 \\ {S}_n=8202.32\end{array}\\ & & \begin{array}{c}\begin{array}{c}8202.32\end{array}=\frac{100\left(1–1.{03}^n\right)}{1–1.03}\\ \left(\frac{–0.03}{100}\right)\begin{array}{c}8202.32\end{array}=\frac{100\left(1–1.{03}^n\right)}{–0.03}\left(\frac{–0.03}{100}\right)\\ –2.4607= 1–1.{03}^n\\ –1=–1\\ –3.4607=–1.{03}^n\\ \log \left(3.4607\right)=\log {\left(1.03\right)}^n\\ \log \left(3.4607\right)=n·\log \left(1.03\right)\\ n=\frac{\log 3.4607}{\log 1.03}=42\end{array}\end{array} \end{gathered}$$

Kristin has been saving for 42 months (3.5 years).

13. Determine the height of the second and third bounce.

$$\begin{gathered} a_1=6 \\ r=0.7 \\ \begin{array}{ccc}{a}_2=a_1{r}^{2–1}& & a_3=a_1{r}^{3–1}\\ a_2=\left(6\right){\left(0.7\right)}^1& & a_2=\left(6\right){\left(0.7\right)}^2\\ a_2=4.2& & a_3=2.94\end{array} \end{gathered}$$

On the second bounce, the ball was 4.2 feet off the ground, and on the third bounce, the ball was 2.94 feet off the ground.

14. After the initial drop, the remaining distances are traveled twice (up and down, or $S_{\mathrm{bounces}}=2\left(S_{\mathrm{heights}}\right)$). How far did the ball travel after four bounces? (Hint: The drop is not counted twice.)

$$\begin{gathered} a_1=6 \\ r=0.7 \\ n=4 \\ {S}_4=\frac{6\left(1–0.7^4\right)}{1–0.7}=15.198 \\ {S}_{\mathrm{distance}}=2\left(15.198\right)–6 \\ {S}_{\mathrm{distance}}=24.396 \end{gathered}$$

The total distance the ball traveled was approximately 24.4 feet.