A pendulum swings 80cm on its first swing, 76cm on its second swing, 72.2cm on i

Yasmin 2021-09-21 Answered
A pendulum swings 80cm on its first swing, 76cm on its second swing, 72.2cm on its third swing, and 68.59cm on its fourth swing. Complete parts a and b below.
a. If the pattern continues, what explicit formula can be used to find the distance of the n-th swing?
\(\displaystyle{a}_{{{n}}}={80}{\left({0.95}\right)}^{{{n}-{1}}}\)
(Simplify your answer. Use integers or decimals for any numbers in the expression.)
b. Use your formula to find the distance of the 10-th swing.
\(\displaystyle{a}_{{{10}}}=?{c}{m}\)
(Type an integer or decimal rounded to two decimal places as needed.)

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Expert Answer

faldduE
Answered 2021-09-22 Author has 14700 answers
Step 1
Given, a pendulum swings 80 cm on its first swing, 76 on its second swing, and 72.2 on its third swing, and 68.69 cm on its fourth swing.
Step 2
Part (a),
A pendulum swings as follows:
First swing, \(\displaystyle{a}_{{{1}}}={80}{c}{m}\)
Second swing, \(\displaystyle{a}_{{{2}}}={76}{c}{m}\)
Third swing, \(\displaystyle{a}_{{{3}}}={72.2}{c}{m}\)
Fourth swing, \(\displaystyle{a}_{{{4}}}={68.59}{c}{m}\)
Common ratio, \(\displaystyle{r}_{{{1}}}={\frac{{{a}_{{{2}}}}}{{{a}_{{{1}}}}}}={\frac{{{76}}}{{{80}}}}={0.95}\)
\(\displaystyle{r}_{{{2}}}={\frac{{{a}_{{{3}}}}}{{{a}_{{{2}}}}}}={\frac{{{72.2}}}{{{76}}}}={0.95}\)
Since the common ratio is same, so the pendulum swings are in geometric sequence.
First term, \(\displaystyle{a}={80}\)
Common ratio, \(\displaystyle{r}={0.95}\)
The n-th term of the geometric sequence is given by
\(\displaystyle\therefore{a}_{{{n}}}={a}{\left({r}\right)}^{{{n}-{1}}};\) where \(\displaystyle{n}=\) number of terms
\(\displaystyle\Rightarrow{a}_{{{n}}}={80}{\left({0.95}\right)}^{{{n}-{1}}}\)
The distance of n-th swings is \(\displaystyle{a}_{{{n}}}={80}{\left({0.95}\right)}^{{{n}-{1}}}\).
Step 3
Part(b),
\(\displaystyle\therefore{a}_{{{n}}}={80}{\left({0.95}\right)}^{{{n}-{1}}}\)
Out \(\displaystyle{n}={10}\), then
\(\displaystyle\Rightarrow{a}_{{{10}}}={80}{\left({0.95}\right)}^{{{10}-{1}}}\)
\(\displaystyle\Rightarrow{a}_{{{10}}}={80}{\left({0.95}\right)}^{{{9}}}\)
\(\displaystyle\Rightarrow{a}_{{{10}}}={50.4199}\)
\(\displaystyle\Rightarrow{a}_{{{10}}}={50.42}{c}{m}\)
The distance of the 10-th swing is \(\displaystyle{a}_{{{10}}}={50.42}{c}{m}\)
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