Geometric Series: Practice

The ball first drops $10$ m. Each rebound goes up and back down, so the bounces total $2(6 + 3.6 + \cdots)$ with $a_1 = 6$, $r = 0.6$.

$10 + 2 \cdot \frac{6}{1 - 0.6} = 10 + 2(15) = 40 \text{ m}$

Use $S = \frac{a_1}{1 - r}$ with $S = 12$ and $a_1 = 3$.

$\frac{3}{1 - r} = 12 \;\Longrightarrow\; 1 - r = \frac{3}{12} = \frac14 \;\Longrightarrow\; r = \frac34$

A two-digit repeat has $a_1 = \frac{48}{100}$ and $r = \frac{1}{100}$.

$0.\overline{48} = \frac{48/100}{1 - 1/100} = \frac{48}{99} = \frac{16}{33}$

Use $S = \frac{a_1}{1 - r}$ with $S = 9$, $a_1 = 6$.

$\frac{6}{1 - r} = 9 \;\Longrightarrow\; 1 - r = \frac23 \;\Longrightarrow\; r = \frac13$

Here $a_1 = 16$ and $r = -\frac12$, and $|r| < 1$, so it converges. Keep the sign.

$S = \frac{16}{1 - (-1/2)} = \frac{16}{3/2} = \frac{32}{3}$

With $a_1 = \frac{81}{100}$ and $r = \frac{1}{100}$,

$0.\overline{81} = \frac{81/100}{1 - 1/100} = \frac{81}{99} = \frac{9}{11}$

The first term (at $k = 1$) is $3\left(\frac14\right)^{0} = 3$, and the ratio is $\frac14$.

$S = \frac{3}{1 - 1/4} = \frac{3}{3/4} = 4$

Each term is $\frac23$ of the last, so $a_1 = \frac23$ and $r = \frac23$.

$S = \frac{2/3}{1 - 2/3} = \frac{2/3}{1/3} = 2$

The terms $3$ through $6$ are $4, 8, 16, 32$. Using $S_6 - S_2$ with $S_n = 2^{n} - 1$,

$S_6 - S_2 = (2^{6} - 1) - (2^{2} - 1) = 63 - 3 = 60$

In cents the daily amounts are $1 + 2 + 4 + \cdots + 2^{29}$, a geometric series with $a_1 = 1$, $r = 2$, $n = 30$.

$S_{30} = \frac{2^{30} - 1}{2 - 1} = 2^{30} - 1 \text{ cents}$

From $S = \frac{a_1}{1 - r}$, solve for the first term: $a_1 = S(1 - r)$.

$a_1 = 20\left(1 - \tfrac25\right) = 20 \cdot \tfrac35 = 12$

With $a_1 = \frac{9}{10}$ and $r = \frac{1}{10}$,

$0.\overline{9} = \frac{9/10}{1 - 1/10} = \frac{9/10}{9/10} = 1$

The repeating decimal $0.\overline{9}$ is exactly equal to $1$, not slightly less.