Geometric Sequences: Additional Practice (Set 1)

Between the 1st and 4th terms there are $4 - 1 = 3$ multiplications by $r$.

$r^{3} = \frac{54}{2} = 27 \Rightarrow r = 3$

A cube root has one real value, so $r = 3$ is settled.

Here $a_1 = 2$ and $r = 4$, so $a_n = 2 \cdot 4^{n-1}$. Set it equal to $128$ and match powers of $4$.

$2 \cdot 4^{n-1} = 128 \Rightarrow 4^{n-1} = 64 = 4^{3} \Rightarrow n = 4$

Isolate the power, take a logarithm, and use the power law, then add $1$ for the position.

$5^{n-1} = 90 \Rightarrow n - 1 = \frac{\log(90)}{\log(5)} \Rightarrow n = 1 + \frac{\log(90)}{\log(5)}$

Use $a_n = a_1 r^{n-1}$ with $n = 6$, so the exponent is $5$.

$a_6 = 3 \cdot (-2)^{5} = 3 \cdot (-32) = -96$

An odd power of the negative ratio is negative.

The middle term is the geometric mean, with $a_2^{2} = a_1 a_3$.

$a_2 = \sqrt{2 \cdot 50} = \sqrt{100} = 10$

The terms are positive, so $a_2 = 10$, not the arithmetic mean $26$.

The terms are $a_n = 5 \cdot 3^{n-1}$. Test terms as they pass $2000$.

$a_6 = 5 \cdot 243 = 1215, \quad a_7 = 5 \cdot 729 = 3645$

The 6th term is still below $2000$, so the first term over $2000$ is $3645$.

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

$r = \frac{a_2}{a_1} = \frac{-6}{3} = -2$

The negative ratio makes the sequence alternate in sign.

Year $4$ uses $n - 1 = 3$ factors of the growth factor $1.2$.

$a_4 = 500 \cdot 1.2^{3} = 500 \cdot 1.728 = 864 \text{ dollars}$

Between the 5th and 9th terms there are $9 - 5 = 4$ multiplications by $r$, so $a_9 = a_5 r^{4}$.

$\frac{a_9}{a_5} = r^{4}$

The area doubles each day, so after $7$ days multiply the start by $2^{7}$.

$1 \cdot 2^{7} = 128 \text{ square meters}$

The terms are $a_n = 8 \cdot \left(\tfrac{1}{2}\right)^{n-1}$. Test terms as they fall below $0.1$.

$a_7 = 0.125, \quad a_8 = 0.0625$

The 7th term is still above $0.1$, so the first term under $0.1$ is $0.0625$.

Over the $4 - 1 = 3$ steps between the terms, isolate the cube of the ratio.

$r^{3} = \frac{12}{96} = \frac{1}{8} \Rightarrow r = \frac{1}{2}$

A cube root has one real value, and $\left(-\tfrac{1}{2}\right)^3 = -\tfrac{1}{8}$, so the ratio is the positive $\tfrac{1}{2}$.