Geometric Series: Free Response

Read off the two defining numbers. The first term is $a_1 = 6$, and the ratio is any term divided by the one before it, $\dfrac{24}{6} = 4$, so $r = 4$. Since $r > 1$, use the form built from positive parts.

$S_5 = \frac{a_1(r^{5} - 1)}{r - 1} = \frac{6(4^{5} - 1)}{4 - 1}.$

Work out the power first, $4^{5} = 1024$, then finish the arithmetic.

$S_5 = \frac{6(1023)}{3} = 2 \times 1023 = 2046.$

Every term of this series is the same number, so the common ratio is $r = 1$. The formula used in part A divides by $r - 1$, which would be dividing by zero here, so that formula cannot be used at all. Instead, the sum is just the repeated term added to itself the number of times it appears.

$S_6 = \underbrace{11 + 11 + \cdots + 11}_{6 \text{ terms}} = 6 \times 11 = 66.$

The formula from part A is $S_n = \dfrac{a_1(r^{n} - 1)}{r - 1}$. Substituting $r = 1$ makes the denominator

$r - 1 = 1 - 1 = 0,$

and dividing by zero is not permitted, so the formula is not merely a bad choice here; it is genuinely undefined. This is not a flaw in the series, only in this one formula's reach.

What replaces it follows directly from what $r = 1$ means: every term equals $a_1$, so summing $n$ of them is just repeated addition, $S_n = n\,a_1$, with no ratio or exponent involved at all.

Divide a term by the one directly before it to find the ratio.

$r = \frac{6}{15} = \frac25.$

Since $\left|\tfrac25\right| < 1$, the infinite-sum formula applies.

$S = \frac{a_1}{1 - r} = \frac{15}{1 - 2/5} = \frac{15}{3/5} = 25.$

$r = \frac{-8}{24} = -\frac13.$

Substitute the negative ratio into $1 - r$ without dropping its sign.

$S = \frac{24}{1 - (-1/3)} = \frac{24}{4/3} = 18.$

Check the condition first: $r = -1$ gives $|r| = 1$, which is not less than $1$, so $\dfrac{a_1}{1 - r}$ does not apply here.

Compute a few partial sums directly.

$S_1 = 7, \quad S_2 = 0, \quad S_3 = 7, \quad S_4 = 0, \ldots$

The total never approaches a single number; it alternates between two values forever, so this series has no finite sum.

Split the decimal at its place values.

$0.\overline{4} = \frac{4}{10} + \frac{4}{100} + \frac{4}{1000} + \cdots$

This is an infinite geometric series with $a_1 = \dfrac{4}{10}$ and $r = \dfrac{1}{10}$; since $|r| < 1$,

$0.\overline{4} = \frac{4/10}{1 - 1/10} = \frac{4/10}{9/10} = \frac49.$

A two-digit block repeats every hundredth, so

$0.\overline{54} = \frac{54}{100} + \frac{54}{10000} + \cdots, \qquad a_1 = \frac{54}{100}, \ r = \frac{1}{100}.$

$0.\overline{54} = \frac{54/100}{1 - 1/100} = \frac{54}{99} = \frac{6}{11}.$

In part A the block is one digit long, so each place is a tenth of the last, and in part B it is two digits long, so each copy is a hundredth of the last.

$r_A = \frac{1}{10}, \qquad r_B = \frac{1}{100}.$

That difference carries straight into $1 - r$:

$1 - r_A = \frac{9}{10}, \qquad 1 - r_B = \frac{99}{100}.$

The length of the repeating block sets how many places over the ratio steps, which is exactly what fixes the power of ten in $1 - r$, and therefore the unreduced denominator, before any reducing takes place.

Every arithmetic step the student wrote is correct: $\dfrac{5}{1-2} = \dfrac{5}{-1} = -5$ really is $-5$. The mistake sits earlier, in reaching for the formula at all before checking what it requires.

The infinite-sum formula $\dfrac{a_1}{1-r}$ was derived under the assumption $|r| < 1$; it is not a universal rule for every ratio. Here

$r = 2 \quad\Longrightarrow\quad |r| = 2 \ge 1,$

so the condition fails, and the formula's output means nothing as a total. The correct verdict is that this series has no finite sum at all, positive or negative.

Add the terms one at a time.

$S_1 = 5, \quad S_2 = 5 + 10 = 15, \quad S_3 = 15 + 20 = 35, \quad S_4 = 35 + 40 = 75.$

Each partial sum is larger than the last, and by an increasing amount, so nothing in this list is heading toward a fixed value, let alone a negative one.

As a bare fraction, $\dfrac{a_1}{1-r}$ can be evaluated for any $r \ne 1$, so it will hand back some real number no matter what ratio is substituted; that is exactly why the student's line of algebra 'worked'. Even a ratio of $r = 3$, which no one would call convergent, still produces a number:

$\frac{a_1}{1 - 3} = \frac{a_1}{-2}.$

What makes a number meaningful as a SUM, though, is the derivation behind the formula, which relies on the later terms of the series shrinking toward $0$ as more are added. That shrinking only happens when $|r| < 1$. When $|r| \ge 1$ the terms do not shrink, so the assumption behind the formula was never true, and its output is disconnected from the series entirely, an artifact of the algebra rather than a report of anything the series is actually doing.

Multiply the sum term by term by $r$.

$rS = a_1 r + a_1 r^{2} + a_1 r^{3} + \cdots.$

Every term on the right is exactly one term of the original sum, shifted over by one position, so this list is the original sum $S$ with its leading $a_1$ removed.

$rS = S - a_1.$

Starting from $rS = S - a_1$, move every term with $S$ to one side.

$S - rS = a_1.$

Factor out $S$ and divide.

$S(1-r) = a_1 \quad\Longrightarrow\quad S = \frac{a_1}{1-r}.$

Look again at the opening line of part A: 'let $S$ denote the sum.' That phrasing quietly assumes such a number exists, ready to be multiplied and subtracted from like any other. Nothing in the multiply-and-subtract step ever questions that assumption; it just carries $S$ through the algebra as if it were already a settled quantity.

Whether $S$ actually is such a number is precisely the question $|r| < 1$ answers. When $|r| < 1$, the running total does settle on one value as more terms are added, so $S$ genuinely exists and the algebra describes something real.

When $|r| \ge 1$, no such settled value exists: the running total either grows without bound or refuses to approach any one number. Naming that non-existent quantity $S$ and doing algebra on it does not summon it into being. With $r = 2$, for instance, the very same steps would claim

$S = \frac{a_1}{1-2} = -a_1,$

a negative number for a sum of terms that only ever grow, which shows the derivation never checked whether its own opening assumption held.