Geometric Sequences: Additional Practice (Set 2)

First find $r$ over the $5 - 2 = 3$ steps between the known terms.

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

Then back up one step from $a_2$, so $a_1 = \tfrac{a_2}{r} = \tfrac{6}{2} = 3$.

Over the $5 - 3 = 2$ steps, $r^{2} = \tfrac{200}{50} = 4$. The first term uses $a_1 = \tfrac{a_3}{r^2}$, and $r^2$ is the same for either sign of $r$.

$a_1 = \frac{50}{4} = 12.5$

The start-of-year balances are geometric with $r = 1.1$, so year $3$ uses $n - 1 = 2$ factors.

$a_3 = 1000 \cdot 1.1^{2} = 1000 \cdot 1.21 = 1210 \text{ dollars}$

First find $r$ over the $5 - 2 = 3$ steps, giving $r^{3} = \tfrac{40}{5} = 8$, so $r = 2$.

$a_7 = a_5 r^{2} = 40 \cdot 4 = 160$

In $20$ years there are $4$ half-lives, so from $a_1 = 240$ take $4$ factors of $\tfrac{1}{2}$.

$240 \cdot \left(\tfrac{1}{2}\right)^{4} = \frac{240}{16} = 15 \text{ mg}$

Use $n = 4$, so the exponent is $3$.

$a_4 = 7 \cdot \left(\tfrac{1}{2}\right)^{3} = \frac{7}{8}$

The terms are $a_n = 100 \cdot \left(\tfrac{1}{2}\right)^{n-1}$. Test terms as they fall below $10$.

$a_4 = 12.5, \quad a_5 = 6.25$

The 4th term is still above $10$, so the first term under $10$ is $6.25$.

Here $a_n = 4 \cdot 2^{n-1}$. Set it equal to $100$ and isolate the power.

$4 \cdot 2^{n-1} = 100 \Rightarrow 2^{n-1} = 25$

Since $25$ is not a power of $2$ (it falls between $16$ and $32$), $100$ is not a term.

After $k$ decades the population is $500 \cdot 3^{k}$. Test values of $k$ near the threshold.

$500 \cdot 3^{3} = 13{,}500, \quad 500 \cdot 3^{4} = 40{,}500$

Three decades fall short, so it first exceeds $40{,}000$ after $4$ decades.

Use $a_n = a_1 r^{n-1}$ with $n = 3$, keeping the negative first term.

$a_3 = -4 \cdot 3^{2} = -4 \cdot 9 = -36$

The ratio is positive, so every term keeps the sign of $a_1$.

Divide a term by the one before it, keeping track of the sign change.

$r = \frac{-10}{5} = -2$

The negative ratio is what makes the signs alternate.

The middle term is the geometric mean of its neighbors, so $x^{2} = 9 \cdot 49$.

$x = \sqrt{9 \cdot 49} = \sqrt{441} = 21$

The value is stated positive, so $x = 21$, not the arithmetic mean $29$.