Geometric Series: Additional Practice (Set 2)

With $a_1 = \frac{6}{10}$ and $r = \frac{1}{10}$,

$0.\overline{6} = \frac{6/10}{1 - 1/10} = \frac{6}{9} = \frac23$

With $a_1 = 1$, $r = 3$, $n = 7$,

$S_7 = \frac{1(3^{7} - 1)}{3 - 1} = \frac{2187 - 1}{2} = 1093$

Each dose adds $100$ mg, and each earlier dose leaves half of itself by the next day, so the long-run just-after-dose amount is $100 + 100\left(\frac12\right) + 100\left(\frac12\right)^{2} + \cdots$ with $a_1 = 100$, $r = \frac12$.

$S = \frac{100}{1 - 1/2} = 200 \text{ mg}$

With $a_1 = 5$, $r = 3$, $n = 4$,

$S_4 = \frac{5(3^{4} - 1)}{3 - 1} = \frac{5(80)}{2} = 200$

Directly, $5 + 15 + 45 + 135 = 200$.

The index starts at $k = 0$, so the first term is $\left(\frac12\right)^{0} = 1$, and the ratio is $\frac12$.

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

The $n$th term is $a_n = 3 \cdot 2^{n-1}$. Set it equal to $768$.

$3 \cdot 2^{n-1} = 768 \;\Longrightarrow\; 2^{n-1} = 256 = 2^{8}$

So $n - 1 = 8$ and $n = 9$.

Each term is $0.9$ of the last, so $a_1 = 1$ and $r = 0.9$, with $|r| < 1$.

$S = \frac{1}{1 - 0.9} = \frac{1}{0.1} = 10$

A ratio close to $1$ makes the sum large, but it is still finite.

An infinite geometric series converges only when $|r| < 1$. The ratio $\frac32$ is larger than $1$.

$\left|\tfrac32\right| = \tfrac32 \ge 1 \;\Rightarrow\; \text{grows without bound}$

The other ratios $\frac12$, $\frac13$, and $-\frac12$ all have size below $1$.

The ratio is $r = \frac{a_2}{a_1} = \frac15$, and $|r| < 1$.

$S = \frac{5}{1 - 1/5} = \frac{5}{4/5} = \frac{25}{4}$

Here $a_1 = \frac52$ and $r = \frac12$, with $|r| < 1$.

$S = \frac{5/2}{1 - 1/2} = \frac{5/2}{1/2} = 5$

The step distances form an infinite geometric series with $a_1 = \frac12$ and $r = \frac12$.

$\frac12 + \frac14 + \frac18 + \cdots = \frac{1/2}{1 - 1/2} = 1$

The infinitely many shrinking steps add to exactly $1$ m, so Achilles does cross the whole meter.

Since $|r| = \frac13 < 1$, the series converges. Keep the sign of the ratio.

$S = \frac{12}{1 - (-1/3)} = \frac{12}{4/3} = 9$