Geometric Sequences: Practice

Set $a_1 r^{n-1} = 320$ and isolate the power, then match to a power of $2$.

$5 \cdot 2^{n-1} = 320 \Rightarrow 2^{n-1} = 64 = 2^{6} \Rightarrow n = 7$

Back up two steps from the third term, dividing by $r$ each time, so $a_1 = \tfrac{a_3}{r^2}$.

$a_1 = \frac{45}{3^{2}} = \frac{45}{9} = 5$

Here $a_n = 3 \cdot 2^{n-1}$. Set it equal to $96$ and match powers of $2$.

$3 \cdot 2^{n-1} = 96 \Rightarrow 2^{n-1} = 32 = 2^{5} \Rightarrow n = 6$

Between the 2nd and 5th terms there are $5 - 2 = 3$ multiplications by $r$.

$r^{3} = \frac{48}{6} = 8 \Rightarrow r = 2$

A cube root has a single real value, so $r = 2$ is settled.

The common ratio is $r = \tfrac{1}{2}$, which lies between $0$ and $1$, so each term is a fraction of the last.

$r = \frac{1}{2}, \quad 0 < r < 1$

The terms decay toward zero without ever reaching it.

The counts double each hour, so multiply the start by $2^3$ after $3$ hours.

$20 \cdot 2^{3} = 20 \cdot 8 = 160$

The two terms are adjacent, so divide the later by the earlier.

$r = \frac{a_4}{a_3} = \frac{36}{12} = 3$

A geometric sequence multiplies by $r$ a total of $n - 1$ times from the first term.

$a_n = a_1 r^{n-1}$

The form $a_1 + (n-1)d$ adds a constant, which describes an arithmetic sequence instead.

Between the 2nd and 5th terms there are $5 - 2 = 3$ factors of $r$, so $a_5 = a_2 r^{3}$.

$a_5 = 10 \cdot 2^{3} = 10 \cdot 8 = 80$

The ratio is a term divided by the one before it, later over earlier.

$r = 15 \div 5 = 3$

Dividing the wrong way, $5 \div 15$, gives the reciprocal, and subtracting gives a difference.

The rebound heights are geometric with $a_1 = 64$ and $r = \tfrac{1}{2}$. The 4th uses $n - 1 = 3$ factors.

$a_4 = 64 \cdot \left(\tfrac{1}{2}\right)^{3} = \frac{64}{8} = 8 \text{ cm}$

The yearly values are geometric with $r = 0.8$, so after $2$ years multiply by $0.8^{2}$.

$20{,}000 \cdot 0.8^{2} = 20{,}000 \cdot 0.64 = 12{,}800 \text{ dollars}$