A Geometric Series is the sum of all the terms in an geometric sequence up to a specific term.
They can be expressed algebraically using a couple of different formulas:
In order to find a particular series value, we can use the explicit formula:
Determine the following of the series \(7-14+28-56+...\):
i. First, we need to find the common ratio in order to determine the \(10\)th term:
\(\textcolor{green}{r} = \cfrac{\textcolor{magenta}{t_n}}{\textcolor{teal}{t_{n-1}}}\)
We can first find the ratio between the second and first terms:
\(\textcolor{green}{r_1} = \cfrac{\textcolor{magenta}{-14}}{\textcolor{teal}{7}}\)
\(\textcolor{green}{r_1 = -2}\)
We can then find the ratio between the third and second terms:
\(\textcolor{green}{r_2} = \cfrac{\textcolor{magenta}{28}}{\textcolor{teal}{-14}}\)
\(\textcolor{green}{r2 = -2}\)
We can now plug all pertinent values into the explicit formula:
\(\textcolor{magenta}{t_n} = \textcolor{red}{a}\textcolor{green}{r}^{\textcolor{blue}{n}-1}\)
\(\textcolor{magenta}{t_{10}} = (\textcolor{red}{7})(\textcolor{green}{-2})^{\textcolor{blue}{10}-1}\)
\(\textcolor{magenta}{t_{10}} = (\textcolor{red}{7})(\textcolor{green}{-2})^9\)
\(\textcolor{magenta}{t_{10}} = (\textcolor{red}{7})(-512)\)
\(\textcolor{magenta}{t_{10} = -3584}\)
Therefore, we can determine the \(10\) term in the series is \(\boldsymbol{-3584}\).
ii. We can find the sum of the first \(10\) terms using the Sum formula:
\(\textcolor{olive}{S_n} = \textcolor{red}{a}\left(\cfrac{1-\textcolor{green}{r}^{\textcolor{blue}{n}}}{1-\textcolor{green}{r}}\right)\)
\(\textcolor{olive}{S_{10}} = \textcolor{red}{7}\left(\cfrac{1-(\textcolor{green}{-2})^{\textcolor{blue}{10}}}{1-(\textcolor{green}{-2})}\right)\)
\(\textcolor{olive}{S_{10}} = \textcolor{red}{7}\left(\cfrac{1-1024}{3}\right)\)
\(\textcolor{olive}{S_{10}} = \textcolor{red}{7}\left(\cfrac{-1023}{3}\right)\)
\(\textcolor{olive}{S_{10}} = \textcolor{red}{7}(-341)\)
\(\textcolor{olive}{S_{10} = -2387}\)
Therefore, we can determine that the sum of all \(25\) terms in this series is \(\boldsymbol{1575}\).
First, we need to identify the common ratio for the series:
\(\textcolor{green}{r} = \cfrac{\textcolor{magenta}{t_n}}{\textcolor{teal}{t_{n-1}}}\)
We can first find the ratio between the second and first terms:
\(\textcolor{green}{r_1} = \cfrac{\textcolor{magenta}{20}}{\textcolor{teal}{5}}\)
\(\textcolor{green}{r_1 = 4}\)
We can then find the ratio between the third and second terms:
\(\textcolor{green}{r_2} = \cfrac{\textcolor{magenta}{80}}{\textcolor{teal}{20}}\)
\(\textcolor{green}{r_2 = 4}\)
Next, we can find the series number through simplfying the explicit formula and using logarithms:
\(\textcolor{magenta}{t_n} = \textcolor{red}{a}\textcolor{green}{r}^{\textcolor{blue}{n}-1}\)
\(\textcolor{magenta}{20480} = \textcolor{red}{5}(\textcolor{green}{4})^{\textcolor{blue}{n}-1}\)
\(\cfrac{\textcolor{magenta}{20480}}{\textcolor{red}{5}} = \cfrac{\cancel{\textcolor{red}{5}}(\textcolor{green}{4})^{\textcolor{blue}{n}-1}}{\cancel{\textcolor{red}{5}}}\)
\(\text{log}(4096) = (\textcolor{blue}{n}-1)(\text{log}(\textcolor{green}{4}))\)
\(\textcolor{blue}{n} - 1 = \cfrac{\text{log}(4096)}{\text{log}(\textcolor{green}{4})}\)
\(\textcolor{blue}{n} - 1 = 6\)
\(\textcolor{blue}{n} = 7\)
We can now plug the series number into the the Sum formula:
\(\textcolor{olive}{S_n} = \textcolor{red}{a}\left(\cfrac{1-\textcolor{green}{r}^{\textcolor{blue}{n}}}{1-\textcolor{green}{r}}\right)\)
\(\textcolor{olive}{S_7} = \textcolor{red}{5}\left(\cfrac{1-(\textcolor{green}{4})^{\textcolor{blue}{7}}}{1-\textcolor{green}{4}}\right)\)
\(\textcolor{olive}{S_7} = \textcolor{red}{5}\left(\cfrac{1-16384}{-3}\right)\)
\(\textcolor{olive}{S_7} = \textcolor{red}{5}\left(\cfrac{-16383}{-3}\right)\)
\(\textcolor{olive}{S_7} = 5(5461)\)
\(\textcolor{olive}{S_7 = 27305}\)
Therefore, we can determine that \(\boldsymbol{S_7 = 27305}\).
In order to determine the total vertical distance, we will need to calulate the distances going up and down separately. First, we need to calculate the initial distance of the ball when it bounces up:
\(a_{\text{Up}} = (\text{bounceratio})(\text{dropheight})\)
\(a_{\text{Up}} = \left(\cfrac{5}{8} \right)(16)\)
\(a_{\text{Up}} = 10\;[\text{m}]\)
We will need to calculate the total vertical distance of the ball going up on its \(6\)th bounce:
Ball Going Up\(\textcolor{olive}{S_n} = \textcolor{red}{a} \left(\cfrac{1-\textcolor{green}{r}^{\textcolor{blue}{n}}}{1-\textcolor{green}{r}}\right)\)
\(\textcolor{olive}{S_6} = \textcolor{red}{10} \left(\cfrac{1-(\textcolor{green}{5/8})^{\textcolor{blue}{6}}}{1-\textcolor{green}{5/8}}\right)\)
\(\textcolor{olive}{S_6} = \textcolor{red}{10} \left(1 - \cfrac{15625}{262144}\right)\left(\cfrac{8}{3}\right)\)
\(\textcolor{olive}{S_6} = \textcolor{red}{10}\left(\cfrac{246519}{262144}\right)\left(\cfrac{8}{3}\right)\)
\(\textcolor{olive}{S_6} = \textcolor{red}{10}\left(\cfrac{1972152}{786432}\right)\)
\(\textcolor{olive}{S_6 \approx 25.1\;[\text{m}]}\)
We will need to calculate the total vertical distance of the ball going down on its \(7\)th bounce:
Ball Going Down\(\textcolor{olive}{S_n} = \textcolor{red}{a}\left(\cfrac{1-\textcolor{green}{r}^{\textcolor{blue}{n}}}{1-\textcolor{green}{r}}\right)\)
\(\textcolor{olive}{S_7} = \textcolor{red}{16}\left(\cfrac{1-(\textcolor{green}{5/8})^{\textcolor{blue}{6}}}{1-\textcolor{green}{5/8}}\right)\)
\(\textcolor{olive}{S_7} = \textcolor{red}{16}\left(1 - \cfrac{15625}{262144}\right)\left(\cfrac{8}{3}\right)\)
\(\textcolor{olive}{S_7} = \textcolor{red}{16}\left(\cfrac{246519}{262144}\right)\left(\cfrac{8}{3}\right)\)
\(\textcolor{olive}{S_7} = \textcolor{red}{16}\left(\cfrac{1972152}{786432}\right)\)
\(\textcolor{olive}{S_7 \approx 40.1\;[\text{m}]}\)
Finally, we can sum both distances to get the total vertical distance of the ball:
\(S_{\text{Total}} = 25.1\;[\text{m}] + 40.1\;[\text{m}]\)
\(S_{\text{Total}} = 65.2\;[\text{m}]\)
Therefore, we can determine that the total vertical distance of the ball when it touches the ground the \(7\)th time is \(\boldsymbol{65.2\;[\textbf{m}]}\).