Geometric Sequences and Series: Free Response

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

$\frac{20}{4} = 5, \qquad \frac{100}{20} = 5, \qquad \frac{500}{100} = 5.$

Every quotient is the same, so the sequence is geometric with $r = 5$.

The explicit formula is

$a_n = 4 \cdot 5^{\,n-1}.$

Reaching the $7$th term from the first uses $7 - 1 = 6$ factors of $r$, not $7$.

$a_7 = 4 \cdot 5^{6} = 4 \cdot 15{,}625 = 62{,}500.$

Check what $4 \cdot 5^{7}$ actually equals in the sequence: it is $a_1 r^{8-1}$, which is the formula for $a_8$, not $a_7$.

$4 \cdot 5^{7} = a_8 = 312{,}500.$

The exponent in $a_n = a_1 r^{\,n-1}$ counts multiplications, and there are only $n - 1$ of those between term $1$ and term $n$: the step from term $1$ to term $2$ is the first multiplication, and the step from term $6$ to term $7$ is the sixth, not the seventh. Using an exponent of $n$ instead always overshoots by exactly one position.

Apply the relation with $k = 2$, $m = 5$:

$\frac{a_5}{a_2} = r^{5-2} \quad\Longrightarrow\quad \frac{-1250}{10} = r^{3} \quad\Longrightarrow\quad r^{3} = -125.$

A cube has exactly one real cube root, so $r = -5$ is the only real solution; there is no second candidate to check.

Back out the first term from $a_2 = a_1 r$:

$a_1 = \frac{a_2}{r} = \frac{10}{-5} = -2.$

The explicit formula is $a_n = -2 \cdot (-5)^{\,n-1}$. For $n = 8$ the exponent is $7$, which is odd, so $(-5)^7$ is negative:

$a_8 = -2 \cdot (-5)^{7} = -2 \cdot (-78{,}125) = 156{,}250.$

Solving for $r$ from two terms always reduces to an equation of the form $r^{\,m-k} = c$, where $m - k$ is the exponent counting the steps between the two known positions and $c = a_m/a_k$.

When that exponent is odd, as it was in part A, the equation has exactly one real solution, since a negative number has exactly one real cube root, just as a positive number does. When the exponent is even instead, $r^{2} = c$ has two real solutions when $c > 0$ and no real solution when $c < 0$; but $c$ is itself an even power of the real ratio $r$ whenever $m-k$ is even, so $c$ can never be negative in this setting, and the two-solution case always applies:

$r^{3} = -125 \;\Rightarrow\; r = -5 \text{ (one root)}, \qquad r^{2} = 25 \;\Rightarrow\; r = \pm 5 \text{ (two roots)}.$

Two terms that are an even number of steps apart therefore cannot, by themselves, tell you the sign of the ratio.

The output triples each week, so $a_1 = 7$ and $r = 3$. Set the explicit formula equal to $5103$:

$7 \cdot 3^{\,n-1} = 5103 \quad\Longrightarrow\quad 3^{\,n-1} = 729.$

Match $729$ to a power of $3$: $729 = 3^{6}$, so $n - 1 = 6$ and $n = 7$.

Since $r = 3 > 1$, use the form of the sum formula built from positive quantities:

$S_7 = \frac{a(r^{7} - 1)}{r - 1} = \frac{7(3^{7} - 1)}{3 - 1} = \frac{7(2187 - 1)}{2} = \frac{7 \cdot 2186}{2} = 7651.$

Substitute $r = 3$, $a_n = 5103$, and $a_1 = 7$ directly:

$S_7 = \frac{3(5103) - 7}{3 - 1} = \frac{15{,}309 - 7}{2} = \frac{15{,}302}{2} = 7651.$

This matches part B exactly. The form is convenient whenever the last term is already known, as it is here from the stem, because it skips recomputing the power $r^{n}$ that the other form of the formula needs.

A weekly rise of $20\%$ means each count is $1 + 0.20 = 1.2$ times the one before it, so $r = 1.2$, not $0.2$. With $a_1 = 300$, the explicit formula is

$a_n = 300 \cdot (1.2)^{\,n-1}.$

The count after $w$ weeks is $300 \cdot (1.2)^{w}$. Require it to exceed $1000$ and isolate the power:

$300(1.2)^{w} > 1000 \quad\Longrightarrow\quad (1.2)^{w} > \tfrac{10}{3}.$

Take logarithms; since $\log(1.2) > 0$, the inequality direction is unchanged:

$w > \frac{\log(10/3)}{\log 1.2} \approx \frac{0.5229}{0.0792} \approx 6.60.$

The smallest whole number above $6.60$ is $w = 7$. Checking both sides confirms it: at $w = 6$ the count is about $300(1.2)^{6} \approx 896$, still under $1000$, and at $w = 7$ it is about $300(1.2)^{7} \approx 1075$, past the target.

The stem sets $a_1$ to be the count with zero weeks of growth behind it. Every additional week of growth advances the term index by one, so the count after $w$ weeks is $a_{w+1}$, not $a_w$:

$a_{w+1} = 300(1.2)^{w}, \qquad w = 7 \;\Rightarrow\; a_{8}.$

The offset is easy to miss because the question is phrased entirely in terms of elapsed weeks, and a week count naturally feels like it should match the term number directly. The stem does spell out the $(w+1)$th-term mapping explicitly, but that habit is strong enough to override it anyway; only deliberately tracking that $a_1$ sits at $w = 0$, one step before $a_2$, keeps the indexing straight.

Every monthly fee equals the one before it, so the ratio of consecutive fees is $9/9 = 1$, a constant ratio, which makes the list genuinely geometric. The general sum formula divides by $1 - r$, and at $r = 1$ that denominator is $0$, so it cannot be used at all here. Instead, add the repeated fee directly:

$S_{12} = 12 \times 9 = 108.$

An $8\%$ yearly rise gives $r = 1.08$, not $0.08$, and $a_1 = 500$ is given directly. Reaching $a_7$ from $a_1$ uses $7 - 1 = 6$ factors of $r$:

$a_7 = 500 \cdot (1.08)^{6} \approx 500 \times 1.586874 \approx 793.44.$

Situation II satisfies both conditions needed for the exponential reading: its ratio $1.08$ is positive and is not equal to $1$, so $f(x) = 500(1.08)^{x}$ is a genuinely rising exponential curve, and the yearly balances are exactly its values at $x = 0, 1, 2, \ldots$

Situation I passes the ratio test for being geometric, since $r = 1$ really is a constant ratio, but it fails the separate condition the exponential reading needs: an exponential function is never allowed to have a base of $1$, because that base is a constant function, not a growth or decay curve.

$1^{x} = 1 \text{ for every } x, \qquad \text{so } f(x) = 1^{x} \text{ is not an exponential curve.}$

So the fee sequence is geometric, but there is no exponential curve for it to be a sample of.