Geometric Series: Additional Practice (Set 1)

Here $a_1 = \frac45$ and $r = \frac15$, with $|r| < 1$.

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

A two-digit repeat has $a_1 = \frac{36}{100}$ and $r = \frac{1}{100}$.

$0.\overline{36} = \frac{36/100}{1 - 1/100} = \frac{36}{99} = \frac{4}{11}$

Here $a_1 = 2$, $r = -2$, $n = 4$. Note $(-2)^{4} = 16$.

$S_4 = \frac{2\left(1 - (-2)^{4}\right)}{1 - (-2)} = \frac{2(1 - 16)}{3} = \frac{-30}{3} = -10$

Adding directly, $2 - 4 + 8 - 16 = -10$.

Use $S = \frac{a_1}{1 - r}$ with $S = 6$, $a_1 = 4$.

$\frac{4}{1 - r} = 6 \;\Longrightarrow\; 1 - r = \frac23 \;\Longrightarrow\; r = \frac13$

The terms are $2, 4, 8, 16, 32$ (first term $2^{1} = 2$, ratio $2$).

$\sum_{k=1}^{5} 2^{k} = \frac{2(2^{5} - 1)}{2 - 1} = 2(31) = 62$

The ball drops $8$ m, then the rebounds (each counted up and down) total $2(4 + 2 + 1 + \cdots)$ with $a_1 = 4$, $r = \frac12$.

$8 + 2 \cdot \frac{4}{1 - 1/2} = 8 + 2(8) = 24 \text{ m}$

A two-digit repeat has $a_1 = \frac{45}{100}$ and $r = \frac{1}{100}$.

$0.\overline{45} = \frac{45/100}{1 - 1/100} = \frac{45}{99} = \frac{5}{11}$

From $S = \frac{a_1}{1 - r}$, solve $a_1 = S(1 - r)$.

$a_1 = 40\left(1 - \tfrac34\right) = 40 \cdot \tfrac14 = 10$

Split off the whole part: $2.\overline{3} = 2 + 0.\overline{3}$, and $0.\overline{3} = \frac{3/10}{1 - 1/10} = \frac13$.

$2.\overline{3} = 2 + \frac13 = \frac73$

The counts are $1 + 2 + 4 + \cdots + 2^{9}$, with $a_1 = 1$, $r = 2$, $n = 10$.

$S_{10} = \frac{2^{10} - 1}{2 - 1} = 1024 - 1 = 1023$

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

$S = \frac{1}{1 - (-1/2)} = \frac{1}{3/2} = \frac23$

With $a_1 = 27$, $r = \frac13$, $n = 3$,

$S_3 = \frac{27\left(1 - (1/3)^{3}\right)}{1 - 1/3} = \frac{27 \cdot \frac{26}{27}}{2/3} = \frac{26}{2/3} = 39$

Directly, $27 + 9 + 3 = 39$.