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

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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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 $$\displaystyle{a}_{{{10}}}={50.42}{c}{m}$$