Geometric Sequences: Free Response

Divide a term by the one directly before it, never subtract.

$r = \frac{15}{5} = 3.$

Check it against another pair: $\frac{45}{15} = 3$ too, so the ratio really is constant.

Reaching the $6$th term from the first uses $n-1=5$ factors of $r$, not $6$.

$a_6 = a_1 r^{6-1} = 5 \cdot 3^{5} = 5 \cdot 243 = 1215.$

Apply the recursive rule once to the carried-forward sixth term:

$a_7 = r\,a_6 = 3(1215) = 3645.$

The explicit formula reaches the same term with $a_7 = a_1 r^{7-1} = a_1 r^{6}$: six factors of $r$, because moving from position $1$ to position $7$ is six steps, and each step contributes exactly one factor of $r$. Using seven factors would overshoot by one extra multiplication.

Apply the relation with $m=1$, $k=3$:

$a_3 = a_1 r^{3-1} \quad \Longrightarrow \quad 18 = 2r^{2}.$

Divide by $2$ and take a square root, keeping BOTH signs since the power is even:

$r^{2} = 9 \quad \Longrightarrow \quad r = \pm 3.$

Multiply $a_1$ by each carried-forward value of $r$.

For $r=3$: $a_2 = 2(3) = 6, \qquad a_3 = 6(3) = 18.$

For $r=-3$: $a_2 = 2(-3) = -6, \qquad a_3 = -6(-3) = 18.$

Both sequences genuinely satisfy $a_1=2$ and $a_3=18$.

The claim says two known terms ALWAYS pin down $r$. Part B built two different, genuine geometric sequences, one with $r=3$ and one with $r=-3$, and both have $a_1=2$ and $a_3=18$ exactly as required. That is a single pair of terms consistent with two different ratios, so the universal claim is false.

What produced the ambiguity is visible in part A: positions $1$ and $3$ are $3-1=2$ steps apart, an EVEN number, so solving for $r$ meant solving an even power, and an even power cannot distinguish a number from its negative:

$(-3)^{2} = 3^{2} = 9.$

Had the two given positions been an odd number of steps apart instead, the equation for $r$ would have been an odd power, which does pin down a single real value.

Set the explicit formula equal to $448$ and isolate the power.

$7 \cdot 2^{n-1} = 448 \quad \Longrightarrow \quad 2^{n-1} = 64.$

Match $64$ to a power of $2$: $64 = 2^{6}$, so $n-1=6$ and $n=7$. That is a positive whole number, so $448$ is genuinely the $7$th term.

$7 \cdot 2^{n-1} = 350 \quad \Longrightarrow \quad 2^{n-1} = 50.$

Since $2^{5}=32$ and $2^{6}=64$, the value $50$ sits strictly between two consecutive powers of $2$, so no whole number $n$ satisfies the equation. Therefore $350$ never appears in this sequence.

Both parts above used the same test: solve $a_1 r^{n-1}=V$ for $n$, and check whether the result is a positive whole number. That is the complete membership condition; there is no separate rule for whether a number LOOKS like it should be in range.

Being between two consecutive terms in size is not enough, because the sequence only visits the specific values $a_1 r^{n-1}$ for whole-number $n$, skipping everything strictly between them. Part B is exactly this: $350$ sits between the actual terms

$a_6 = 7 \cdot 2^{5} = 224 \qquad \text{and} \qquad a_7 = 7 \cdot 2^{6} = 448,$

well inside the sequence's range, yet $2^{n-1}=50$ has no whole-number solution, so $350$ is skipped. Only $448$, at the position the equation actually lands on, is a genuine term.

Test each list both ways.

List I: differences $20-10=10$, $30-20=10$, $40-30=10$ are constant, so it is arithmetic with $d=10$. Its ratios $\tfrac{20}{10}=2$, $\tfrac{30}{20}=1.5$, $\tfrac{40}{30}=\tfrac{4}{3}$ are not constant, so it is not geometric.

List II: ratios $\frac{20}{10}=2, \quad \frac{40}{20}=2, \quad \frac{80}{40}=2$ are constant, so it is geometric with $r=2$.

List III: ratios $\frac{-20}{10}=-2, \quad \frac{40}{-20}=-2, \quad \frac{-80}{40}=-2$ are constant, so it is geometric with $r=-2$.

Multiply by the carried-forward ratio one step at a time.

$-80 \times (-2) = 160, \qquad 160 \times (-2) = -320.$

The signs run $+,-,+,-,+,-$, alternating every step, while the sizes $10,20,40,80,160,320$ keep growing since $|r|=2>1$. A negative ratio does not make a sequence decrease; it makes it alternate.

Every step of a geometric sequence multiplies the previous term by $r$. When $r$ is negative, that single multiplication flips the sign of the term regardless of how large or small $|r|$ is, so alternation is purely a consequence of the SIGN of $r$, not its size.

The SIZE of $r$ decides something different: whether the terms grow or shrink. Compare the two magnitudes directly:

$|-2| = 2 > 1, \qquad |-0.5| = 0.5 < 1.$

With $r=-2$, besides alternating, the sizes grow without bound. If instead $r=-0.5$, the terms would still alternate in sign step to step, but their sizes would shrink toward $0$ instead of growing, the way a positive ratio between $0$ and $1$ decays. Sign and size are two separate effects of $r$, and a negative ratio never by itself means decreasing.

The dose sets $a_1=160$, and each hour scales the previous amount by the fixed factor $0.75$, so $r=0.75$. Reaching hour $4$ from hour $1$ uses $n-1=3$ factors of $r$:

$a_4 = a_1 r^{4-1} = 160(0.75)^{3} = 160(0.421875) = 67.5 \text{ mg}.$

Set up the inequality and isolate the power.

$160(0.75)^{n-1} < 30 \quad \Longrightarrow \quad (0.75)^{n-1} < 0.1875.$

Take a logarithm of both sides. Because $\log(0.75)$ is negative, dividing by it reverses the inequality:

$n-1 > \frac{\log(0.1875)}{\log(0.75)} \approx 5.82.$

So $n-1$ must be at least $6$, giving $n=7$. Checking confirms it: at $n-1=5$ the amount is about $37.97$ mg, still above $30$, while at $n-1=6$ it is about $28.48$ mg, the first value below $30$.

Every term after the first is the term before it times $r=0.75$:

$a_n = 0.75\,a_{n-1}.$

Since $0.75 \ne 0$, multiplying a nonzero amount by it always produces another nonzero amount, so by the same reasoning that applies to any geometric sequence with $0<r<1$, no term of this model is ever exactly $0$; the amounts only shrink toward $0$ without ever landing on it.

Physically, this means the model guarantees the remaining amount will eventually fall below any threshold you pick, which is exactly what part B found for $30$ mg. It does not mean the drug is claimed to remain forever in some literal physical sense; a decaying geometric model is an idealization of clearance, and treating its never-quite-zero property as a literal physical claim would be reading more into the model than it was built to say.