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Sequences and Series: Chapter Test

20 multiple-choice questions and 10 free-response questions, drawn from across the chapter and mixed together.

Free response

10 questions in parts, 96 points in total. Work them out on paper. There are no hints here: reveal each question's answer, worked solution, and rubric when you are ready to mark that one.

Free response · work it on paper

Question 1 of 10

1. A sequence that falls by a fixed step . 9 points. Question 1 of 10.

A sequence begins $50,\ 44,\ 38,\ 32,\ \ldots$ and every term after the first decreases by the same amount.
1. Part A.
  State $a_1$ and the common difference $d$ , then use the explicit formula to find $a_{12}$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Which term of the sequence equals $-100$ ? If no term equals $-100$ , say so and explain how you know.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Using the single step from $a_1$ to $a_2$ , explain why the explicit formula reaches the $n$ th term with $n-1$ steps of $d$ rather than $n$ .
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
The answer

Part A

$a_1 = 50$ , $d = -6$ , and $a_{12} = -16$ .

Part B

$n = 26$ : the $26$ th term equals $-100$ , because solving gives a positive whole number.

Part C

Going from $a_1$ to $a_2$ is a single step of $d$ , and $a_1$ itself sits at position $1$ with zero steps taken, so reaching position $n$ takes $n-1$ steps. Multiplying by $n$ would count one step too many, overshooting every term after the first.

Worked solution

Part A

The common difference is $d = 44 - 50 = -6$ , and reaching the twelfth term takes $n - 1 = 11$ steps.

$a_{12} = 50 + (12 - 1)(-6) = 50 - 66 = -16$

Part B

Set the explicit formula equal to $-100$ and solve for $n$ .

$50 + (n - 1)(-6) = -100 \quad\Longrightarrow\quad (n - 1)(-6) = -150 \quad\Longrightarrow\quad n = 26$

Since $26$ is a positive whole number, $-100$ is genuinely a term.

Part C

From $a_1 = 50$ to $a_2 = 44$ is one application of the step $d = -6$ , not two, and $a_1$ needed zero steps because it is where the counting starts.

$a_2 = a_1 + d, \qquad a_n = a_1 + (n - 1)d$

Following that count, position $n$ is reached after exactly $n - 1$ steps, one fewer than the position number, which is why the formula multiplies $d$ by $n - 1$ .

In one line

For $50, 44, 38, 32, \ldots$ : $a_1 = 50$ , $d = -6$ , and $a_{12} = -16$ ; the value $-100$ is the $26$ th term, since solving $a_n = -100$ gives the positive whole number $n = 26$ ; and the formula uses $n-1$ steps because $a_1$ takes zero steps and each later term adds exactly one.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Finds $d$ as a later term minus the earlier one, and applies $n - 1 = 11$ steps rather than $12$ . . Worth 2 points.

Reports the computed value as the twelfth term itself, not its position. . Worth 1 point.

Part B 3 points

Sets $a_n = -100$ and solves the resulting equation for $n$ . . Worth 2 points.

Bases the membership verdict on whether $n$ is a positive whole number. . Worth 1 point.

Part C 3 points

Shows the transition from $a_1$ to $a_2$ is one step, and generalizes that position $n$ takes $n-1$ steps, not $n$ . . Worth 3 points. needs an explanation, not just an answer
2. Two terms, and how many ratios fit them . 9 points. Question 2 of 10.

A geometric sequence has $a_2 = 12$ and $a_4 = 3$ .
1. Part A.
  Using $a_k = a_m r^{k-m}$ , find every possible value of the common ratio $r$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  For each ratio from part A, find the first term $a_1$ .
  
  Carry your own answer forward Use whichever ratio or ratios you found in part A, even if they differ from the expected ones.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Explain what feature of positions $2$ and $4$ made the ratio ambiguous here, and describe how the two given positions would have to differ for the ratio to be pinned down to a single value.
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
The answer

Part A

$r = \dfrac{1}{2}$ or $r = -\dfrac{1}{2}$ .

Part B

$a_1 = 24$ when $r = \dfrac{1}{2}$ , and $a_1 = -24$ when $r = -\dfrac{1}{2}$ .

Part C

Positions $2$ and $4$ are an even number of steps apart, so $r$ satisfies an even power, $r^2 = \frac{1}{4}$ , which cannot distinguish a value from its negative. Had the two given positions been an odd number of steps apart, the equation for $r$ would be an odd power, which fixes a single real value.

Worked solution

Part A

Positions $2$ and $4$ are $2$ steps apart, so $a_4 = a_2 r^{2}$ . The power is even, so keep both signs of the root.

$3 = 12 r^{2} \quad\Longrightarrow\quad r^{2} = \frac{1}{4} \quad\Longrightarrow\quad r = \pm\frac{1}{2}$

Part B

The first term is one step before $a_2$ , so $a_1 = \dfrac{a_2}{r}$ . Divide $12$ by each carried-forward ratio.

$a_1 = \frac{12}{1/2} = 24, \qquad a_1 = \frac{12}{-1/2} = -24$

Part C

The gap is $4 - 2 = 2$ , an even number, so solving for $r$ meant solving an even power, and an even power hides the sign:

$\left(\tfrac{1}{2}\right)^{2} = \left(-\tfrac{1}{2}\right)^{2} = \frac{1}{4}$

An odd distance would give an odd power, which takes a unique real root, so the two terms would then determine $r$ completely.

In one line

From $a_2 = 12$ and $a_4 = 3$ , the ratio solves $r^2 = \frac{1}{4}$ , so $r = \frac{1}{2}$ or $r = -\frac{1}{2}$ , giving $a_1 = 24$ or $a_1 = -24$ . The ambiguity is because positions $2$ and $4$ are an even number of steps apart; an odd distance would fix $r$ to a single value.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Sets up $a_4 = a_2 r^{2}$ with the correct exponent $4 - 2$ and isolates $r^{2}$ . . Worth 2 points.

Reports both values of $r$ , not just the positive one. . Worth 1 point.

Part B 3 points

Recovers $a_1$ from $a_2$ by dividing by each carried-forward ratio. . Worth 2 points.

Pairs each first term with the ratio that produced it, keeping the two cases distinct. . Worth 1 point.

Part C 3 points

Identifies the even distance as the source of the two solutions, tying it to the even power hiding the sign. . Worth 2 points. needs an explanation, not just an answer

States that an odd distance would pin $r$ to a single value. . Worth 1 point.
3. One series totalled two ways, and a stretch in the middle . 10 points. Question 3 of 10.

An arithmetic sequence has first term $a_1 = 9$ and common difference $d = 7$ .
1. Part A.
  Find $S_{20}$ , the sum of the first $20$ terms.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Find the sum of the $8$ th through $20$ th terms, inclusive.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Explain why the number of terms from position $m$ to position $n$ , inclusive, is $n - m + 1$ rather than $n - m$ , and why totalling such a range still uses the ordinary arithmetic-series formula.
  
  Explain why it works A sentence or two. Reasons, not steps. 3 points
The answer

Part A

$S_{20} = 1510$ .

Part B

$1300$ ; the range has $20 - 8 + 1 = 13$ terms.

Part C

Both endpoints are counted, so plain subtraction $n - m$ counts only the gaps and leaves one endpoint out; adding it back gives $n - m + 1$ (here $20 - 8 + 1 = 13$ ). The range is itself an arithmetic series with its own first and last terms, so the same average-of-the-ends formula applies to it.

Worked solution

Part A

The twentieth term is $a_{20} = 9 + 19(7) = 142$ , so use the average-of-the-ends form.

$S_{20} = \frac{20(9 + 142)}{2} = 10 \cdot 151 = 1510$

Part B

The range is its own arithmetic series with $a_8 = 9 + 7(7) = 58$ , $a_{20} = 142$ , and $20 - 8 + 1 = 13$ terms.

$S = \frac{13(58 + 142)}{2} = 13 \cdot 100 = 1300$

Part C

Subtracting the positions, $20 - 8$ , counts only the gaps between them, not the two endpoint terms themselves. Both position $8$ and position $20$ belong to the range, so one more term than the gap is included.

$20 - 8 + 1 = 13$

Those $13$ consecutive terms are equally spaced, so they form an arithmetic series in their own right, and the same formula totals them.

In one line

With $a_1 = 9$ and $d = 7$ : $S_{20} = 1510$ , and the $8$ th through $20$ th terms total $1300$ across $20 - 8 + 1 = 13$ terms. The count is $13$ because both endpoints are included, and the range totals with the ordinary formula because it is itself an arithmetic series.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Finds the last term and applies a correct sum formula with $n = 20$ . . Worth 2 points.

Reports the result as the total of all twenty terms, not a single term. . Worth 1 point.

Part B 4 points

Counts the terms in the range inclusively, using $n - m + 1$ rather than $n - m$ . . Worth 2 points.

Finds the range's own first and last term and totals it correctly. . Worth 1 point.

Reports the value as the total of the whole range, not a single term. . Worth 1 point.

Part C 3 points

Explains that both endpoints are counted, so plain subtraction leaves one out, giving $n - m + 1$ . . Worth 2 points. needs an explanation, not just an answer

States that the range is itself an arithmetic series, so the ordinary formula applies. . Worth 1 point.
4. A telescoping evaluation, checked line by line . 10 points. Question 4 of 10.

Here is a proposed evaluation of $\displaystyle\sum_{k=1}^{n}\frac{1}{(k+1)(k+3)}$ , given line by line, each line claimed to follow from the one directly above it.

Line 1: split each term, $\dfrac{1}{(k+1)(k+3)} = \dfrac{1}{2}\left(\dfrac{1}{k+1} - \dfrac{1}{k+3}\right)$ .

Line 2: expand the sum, $\dfrac{1}{2}\left[\left(\dfrac{1}{2} - \dfrac{1}{4}\right) + \left(\dfrac{1}{3} - \dfrac{1}{5}\right) + \left(\dfrac{1}{4} - \dfrac{1}{6}\right) + \cdots + \left(\dfrac{1}{n+1} - \dfrac{1}{n+3}\right)\right]$ .

Line 3: cancel the interior, $\dfrac{1}{2}\left(\dfrac{1}{2} - \dfrac{1}{n+3}\right)$ .

Line 4: simplify, $\dfrac{1}{4} - \dfrac{1}{2(n+3)}$ .
1. Part A.
  Identify the first line that does not validly follow from the line directly above it, and state exactly what is wrong.
  
  Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 3 points
2. Part B.
  Write the correct closed form for $\displaystyle\sum_{k=1}^{n}\frac{1}{(k+1)(k+3)}$ .
  
  Carry your own answer forward Apply the corrected survivor pairing you identified in part A.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  Explain, in general and not just for this sum, why a gap-2 telescope leaves two survivors at each end while a gap-1 telescope like $\dfrac{1}{k(k+1)}$ leaves only one.
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
The answer

Part A

Line 3. With a gap of $2$ , each negative piece cancels its partner two brackets later, so two pieces survive at each end: $\frac{1}{2}$ and $\frac{1}{3}$ at the front, $\frac{1}{n+2}$ and $\frac{1}{n+3}$ at the back, not one of each.

Part B

$\dfrac{5}{12} - \dfrac{1}{2}\left(\dfrac{1}{n+2} + \dfrac{1}{n+3}\right)$ .

Part C

In a gap-1 telescope the negative piece of a bracket is cancelled by the very next bracket, so all but the first and last cancel. In a gap-2 telescope the partner is two brackets away, so the first two positive pieces and the last two negative pieces never reach a partner and survive.

Worked solution

Part A

Lines 1 and 2 are sound: the split is the correct gap-2 split, and the expansion from $k = 1$ is faithful. Line 3 is the first that fails. Because the factors differ by $2$ , the negative piece $-\frac{1}{4}$ from the $k = 1$ bracket is not cancelled until the $k = 3$ bracket, two later.

$\text{front survivors: } \tfrac{1}{2}, \tfrac{1}{3}; \qquad \text{back survivors: } -\tfrac{1}{n+2}, -\tfrac{1}{n+3}$

So two pieces survive at each end, and Line 3 kept only one of each.

Part B

Keep both front survivors and both back survivors, with the factor $\frac{1}{2}$ out front.

$\frac{1}{2}\left(\frac{1}{2} + \frac{1}{3} - \frac{1}{n+2} - \frac{1}{n+3}\right) = \frac{5}{12} - \frac{1}{2}\left(\frac{1}{n+2} + \frac{1}{n+3}\right)$

Part C

The gap between the factors sets how many brackets apart a term meets its cancelling partner.

$\text{gap } 1: -\tfrac{1}{k+1} \text{ meets } +\tfrac{1}{k+1} \text{ one bracket later}; \qquad \text{gap } 2: -\tfrac{1}{k+2} \text{ meets } +\tfrac{1}{k+2} \text{ two brackets later}$

With the delay of two, the first two positives and the last two negatives run off the end of the range with no partner, so two survive at each end.

In one line

Line 3 is the first invalid line: a gap-2 telescope leaves two survivors at each end, $\frac{1}{2}$ and $\frac{1}{3}$ at the front and $\frac{1}{n+2}$ , $\frac{1}{n+3}$ at the back, so the correct closed form is $\frac{5}{12} - \frac{1}{2}\left(\frac{1}{n+2} + \frac{1}{n+3}\right)$ . Two survive because the factors differ by $2$ , delaying each cancellation by two brackets.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Names the first invalid line and clears the lines before it as sound. . Worth 1 point.

Explains, from the gap-2 survivor rule, why two pieces survive at each end. . Worth 2 points. needs an explanation, not just an answer

Part B 4 points

Combines both front survivors and both back survivors, not one of each. . Worth 2 points.

Simplifies the front pieces and reports the full corrected closed form. . Worth 2 points.

Part C 3 points

Ties the survivor count to how many brackets apart a term meets its cancelling partner, contrasting gap-1 with gap-2. . Worth 2 points. needs an explanation, not just an answer

States the reasoning holds for any gap-2 telescope, not just this one. . Worth 1 point.
5. A finite geometric total, and the case the formula cannot touch . 9 points. Question 5 of 10.

Two finite series are given. Series P is $4 + 12 + 36 + 108 + 324$ . Series Q is seven copies of the same number, $13 + 13 + 13 + 13 + 13 + 13 + 13$ .
1. Part A.
  Find the sum of Series P using the finite geometric-sum formula.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Find the sum of Series Q, and say in one line why the formula from part A cannot be used on it.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Explain why $r = 1$ is the one ratio the finite formula $S_n = \dfrac{a_1(r^n - 1)}{r - 1}$ cannot handle, and state the rule that replaces it.
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
The answer

Part A

$S_5 = 484$ .

Part B

$S_7 = 91$ ; the part A formula divides by $r - 1 = 0$ when $r = 1$ , so it is undefined, and $S_n = n\,a_1$ is used instead.

Part C

At $r = 1$ the denominator $r - 1$ is $0$ , so the fraction is undefined, not merely awkward. When $r = 1$ every term equals $a_1$ , so the total is plain repeated addition, $S_n = n\,a_1$ , with no ratio or exponent involved.

Worked solution

Part A

Series P has $a_1 = 4$ , $r = \frac{12}{4} = 3$ , and $5$ terms, so use the form that stays positive for $r > 1$ .

$S_5 = \frac{4(3^{5} - 1)}{3 - 1} = \frac{4(243 - 1)}{2} = 484$

Part B

Every term of Series Q is equal, so $r = 1$ , and the formula from part A divides by $r - 1 = 0$ . Instead the sum is repeated addition.

$S_7 = 7 \times 13 = 91$

Part C

Substituting $r = 1$ into the formula makes the denominator vanish.

$r - 1 = 1 - 1 = 0$

Dividing by zero is not permitted, so the formula is genuinely undefined there, not a bad choice. What replaces it follows from what $r = 1$ means: every term is $a_1$ , so summing $n$ of them gives $S_n = n\,a_1$ .

In one line

Series P sums to $S_5 = 484$ by the finite geometric-sum formula with $r = 3$ . Series Q has $r = 1$ , where that formula divides by $r - 1 = 0$ and is undefined, so its total is $S_7 = 7 \times 13 = 91$ by the replacement rule $S_n = n\,a_1$ .

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Reads off $a_1$ and $r$ and applies the finite geometric-sum formula with $n = 5$ . . Worth 2 points.

Reports the result as the total of all five terms, not one of them. . Worth 1 point.

Part B 3 points

Recognizes $r = 1$ and totals with $S_n = n\,a_1$ rather than the ratio-based formula. . Worth 2 points.

Reports the total as the term added to itself seven times. . Worth 1 point.

Part C 3 points

Ties the breakdown to the denominator $r - 1$ being zero at $r = 1$ , making the formula undefined. . Worth 2 points. needs an explanation, not just an answer

States the replacement rule $S_n = n\,a_1$ for a constant series. . Worth 1 point.
6. Two savings plans, two kinds of growth . 10 points. Question 6 of 10.

Two people start saving. Avery deposits $10$ dollars on day $1$ and each day after that deposits $6$ dollars more than the day before. Blair deposits $3$ dollars on day $1$ and each day after that deposits triple the previous day's deposit. These describe the daily deposits, not the running totals.
1. Part A.
  Identify each person's daily deposits as an arithmetic or a geometric sequence, and give the first term together with the common difference or common ratio.
  
  Model the situation Name your unknown first, then write every other quantity in terms of that one letter. 3 points
2. Part B.
  Find each person's deposit on day $6$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Explain how the two deposit lists alone let you tell which plan is arithmetic and which is geometric, and describe what each person's daily deposits do in the long run.
  
  Compare the two methods Say what each one costs you, and when you would reach for it. 3 points
The answer

Part A

Avery's deposits are arithmetic: $a_1 = 10$ dollars, $d = 6$ dollars. Blair's deposits are geometric: $a_1 = 3$ dollars, $r = 3$ .

Part B

Avery deposits $40$ dollars on day $6$ ; Blair deposits $729$ dollars on day $6$ .

Part C

Divide, do not subtract: Avery's consecutive deposits share a constant difference of $6$ dollars but not a constant ratio, so arithmetic; Blair's share a constant ratio of $3$ but not a constant difference, so geometric. Avery's deposits grow by a steady amount each day; Blair's grow by an ever-larger amount, without bound.

Worked solution

Part A

Avery's deposits change by a fixed added amount each day, a constant difference, so they are arithmetic with $a_1 = 10$ and $d = 6$ . Blair's deposits change by a fixed multiplier each day, a constant ratio, so they are geometric with $a_1 = 3$ and $r = 3$ .

$\text{Avery: } d = 6, \qquad \text{Blair: } r = 3$

Part B

Day $6$ is $n - 1 = 5$ steps or factors from day $1$ .

$\text{Avery: } 10 + 5(6) = 40, \qquad \text{Blair: } 3 \cdot 3^{5} = 729$

Part C

Test each list both ways. Avery's deposits differ by $6$ every day, but their ratios are not constant ( $\frac{16}{10} \ne \frac{22}{16}$ ), so arithmetic. Blair's deposits have ratio $3$ every day, but their differences are not constant, so geometric.

$\text{arithmetic: steady increase} \qquad \text{geometric: increase multiplies without bound}$

Avery's deposits rise by the same $6$ dollars forever; Blair's each jump is triple the last, so the amounts grow faster and faster.

In one line

Avery's deposits are arithmetic ( $a_1 = 10$ dollars, $d = 6$ dollars) and Blair's are geometric ( $a_1 = 3$ dollars, $r = 3$ ), so on day $6$ Avery deposits $40$ dollars and Blair deposits $729$ dollars. Constant difference marks the arithmetic plan and constant ratio the geometric one; Avery's deposits grow steadily while Blair's grow without bound.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Classifies Avery's deposits as arithmetic and Blair's as geometric by the constant-difference versus constant-ratio test. . Worth 2 points.

States the first term and the common difference or ratio for each plan. . Worth 1 point.

Part B 4 points

Finds Avery's day-6 deposit with $n - 1 = 5$ steps of the common difference. . Worth 1 point.

Finds Blair's day-6 deposit with $n - 1 = 5$ factors of the common ratio. . Worth 2 points.

Reports both amounts with the unit, dollars. . Worth 1 point.

Part C 3 points

Distinguishes the two by testing for a constant difference versus a constant ratio, dividing rather than subtracting for the geometric one. . Worth 2 points. needs an explanation, not just an answer

Describes the long-run behavior of each: steady increase versus increase that multiplies without bound. . Worth 1 point.
7. An infinite total, and the condition it depends on . 9 points. Question 7 of 10.

Consider the infinite geometric series $\dfrac{5}{2} + \dfrac{5}{4} + \dfrac{5}{8} + \cdots$ , and, separately, the repeating decimal $0.\overline{72}$ .
1. Part A.
  Find the sum of $\dfrac{5}{2} + \dfrac{5}{4} + \dfrac{5}{8} + \cdots$ , confirming the convergence condition before you sum.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Express the repeating decimal $0.\overline{72}$ as an exact fraction in lowest terms by treating it as an infinite geometric series.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
3. Part C.
  Explain why the infinite geometric-sum formula requires $\left|r\right| < 1$ , and what its output would mean if it were applied to a series with $\left|r\right| \ge 1$ .
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
The answer

Part A

$\left|r\right| = \dfrac{1}{2} < 1$ , so the sum exists and equals $5$ .

Part B

$0.\overline{72} = \dfrac{72}{99} = \dfrac{8}{11}$ .

Part C

The formula comes from the running total settling as terms are added, which happens only when the terms shrink, i.e. when $|r| < 1$ . For $|r| > 1$ or $r = -1$ the terms do not shrink, so the fraction evaluates to a number that is a sum of nothing; at $r = 1$ it is undefined ( $1 - r = 0$ ).

Worked solution

Part A

The ratio is $r = \frac{5/4}{5/2} = \frac{1}{2}$ , and $\left|\frac{1}{2}\right| < 1$ , so the infinite-sum formula applies.

$S = \frac{a_1}{1 - r} = \frac{5/2}{1 - 1/2} = \frac{5/2}{1/2} = 5$

Part B

A two-digit block repeats every hundredth, so $a_1 = \frac{72}{100}$ and $r = \frac{1}{100}$ .

$0.\overline{72} = \frac{72/100}{1 - 1/100} = \frac{72}{99} = \frac{8}{11}$

Part C

The derivation depends on the later terms shrinking toward $0$ as more are added, so that the running total approaches one value. That shrinking happens exactly when $|r| < 1$ .

$|r| \ge 1 \quad\Longrightarrow\quad \text{the terms do not shrink, so the total grows or oscillates}$

The bare fraction $\frac{a_1}{1-r}$ can be evaluated for almost any $r$ , but outside $|r| < 1$ its value is an artifact of the algebra, not a total the series reaches.

In one line

The series $\frac{5}{2} + \frac{5}{4} + \cdots$ has $r = \frac{1}{2}$ , so $|r| < 1$ and it sums to $5$ ; the repeating decimal $0.\overline{72} = \frac{72}{99} = \frac{8}{11}$ . The formula needs $|r| < 1$ because only then do the terms shrink and the running total settle; for $|r| > 1$ or $r = -1$ it still returns a number, but not a sum of anything, and at $r = 1$ it is undefined.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Finds $r$ by dividing and checks $|r| < 1$ before applying the formula. . Worth 2 points.

Reports the result as the settled total of the whole infinite series. . Worth 1 point.

Part B 3 points

Reads off the first term and ratio from the place value of the repeating block. . Worth 1 point.

Sums to a fraction and reduces it to lowest terms. . Worth 2 points.

Part C 3 points

Ties the $|r| < 1$ requirement to the terms shrinking so the running total settles. . Worth 2 points. needs an explanation, not just an answer

States that the formula still outputs a number for $|r| \ge 1$ , but it is not a genuine sum. . Worth 1 point.
8. Two telescoping tails, and which one settles . 9 points. Question 8 of 10.

Two infinite telescoping sums are offered: $\displaystyle\sum_{k=1}^{\infty}\left(\frac{1}{k+3} - \frac{1}{k+4}\right)$ and $\displaystyle\sum_{k=1}^{\infty}\left(\sqrt{k+1} - \sqrt{k}\right)$ .
1. Part A.
  Find the closed form of the first sum's first $n$ terms, and state its value as $n$ grows without bound.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Find the closed form of the second sum's first $n$ terms, and state what happens as $n$ grows without bound.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
3. Part C.
  Both sums telescope to two survivors. Explain what actually decides whether an infinite telescoping sum has a finite value, and use it to say, for each of these two sums, whether it settles.
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
The answer

Part A

$\dfrac{1}{4} - \dfrac{1}{n+4}$ , which approaches $\dfrac{1}{4}$ as $n$ grows.

Part B

$\sqrt{n+1} - 1$ , which grows without bound as $n$ grows.

Part C

What decides it is whether the surviving tail term shrinks to zero, not whether the sum telescopes. The first's tail $\frac{1}{n+4}$ shrinks to zero, so it settles at $\frac{1}{4}$ ; the second's tail $\sqrt{n+1}$ grows, so it never settles. Telescoping alone is not enough.

Worked solution

Part A

With $b_k = \frac{1}{k+3}$ , the sum is a chain of $b_k - b_{k+1}$ that collapses to $b_1 - b_{n+1}$ .

$\sum_{k=1}^{n}\left(\frac{1}{k+3} - \frac{1}{k+4}\right) = \frac{1}{4} - \frac{1}{n+4} \;\longrightarrow\; \frac{1}{4}$

Part B

This sum is already in mirror form $b_{k+1} - b_k$ with $b_k = \sqrt{k}$ , so it collapses to the last value minus the first.

$\sum_{k=1}^{n}\left(\sqrt{k+1} - \sqrt{k}\right) = \sqrt{n+1} - 1 \;\longrightarrow\; \infty$

Part C

Both collapse to a fixed front piece and a moving tail. The value settles precisely when that tail shrinks to zero.

$\frac{1}{4} - \frac{1}{n+4} \to \frac{1}{4} \quad (\text{tail} \to 0), \qquad \sqrt{n+1} - 1 \to \infty \quad (\text{tail} \to \infty)$

So the first is ruled in and the second ruled out; the deciding feature is the behavior of the surviving tail, not the fact of telescoping.

In one line

The first sum collapses to $\frac{1}{4} - \frac{1}{n+4}$ , whose tail shrinks to zero, so it settles at $\frac{1}{4}$ ; the second collapses to $\sqrt{n+1} - 1$ , whose tail grows without bound, so it never settles. What decides an infinite telescoping sum is whether its surviving tail shrinks to zero, not that it telescopes.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Collapses the first sum to its correct closed form. . Worth 2 points.

States that the tail $\frac{1}{n+4}$ shrinks to zero, so the form approaches $\frac{1}{4}$ . . Worth 1 point.

Part B 3 points

Collapses the second sum to its correct closed form. . Worth 2 points.

States that $\sqrt{n+1}$ grows without bound, so the form has no limiting value. . Worth 1 point.

Part C 3 points

Identifies the shrinking of the surviving tail as the deciding feature, not the telescoping itself. . Worth 2 points. needs an explanation, not just an answer

Applies it to rule the first sum in and the second out. . Worth 1 point.
9. A bouncing ball and the distance it travels . 10 points. Question 9 of 10.

A ball is dropped from a height of $9$ feet. Each bounce rises to $\dfrac{2}{3}$ of the height of the previous bounce, so the first bounce rises to $\dfrac{2}{3}$ of the drop height.
1. Part A.
  Treating the bounce heights as a geometric sequence, find the height the ball rises to on its third bounce.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Find the total vertical distance the ball travels before coming to rest.
  
  Carry your own answer forward Use the first term and ratio of the bounce-height sequence you set up in part A to build the up-and-down distances.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  The ball bounces infinitely many times in this model, yet the total distance is finite. Explain how infinitely many bounces can add to a finite distance, and say what the condition $\left|r\right| < 1$ has to do with it.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 3 points
The answer

Part A

The third bounce rises to $\dfrac{8}{3}$ feet.

Part B

$45$ feet: the $9$ -foot drop, plus twice the infinite sum $6 + 4 + \dfrac{8}{3} + \cdots = 18$ for the up-and-down of every bounce.

Part C

Each bounce is $\frac{2}{3}$ of the last, so the rise heights form a geometric series with $|r| = \frac{2}{3} < 1$ ; the terms shrink fast enough that the running total settles instead of growing. Infinitely many terms add to a finite value precisely because $|r| < 1$ makes them shrink toward zero.

Worked solution

Part A

The first bounce rises to $9 \cdot \frac{2}{3} = 6$ feet, so the bounce heights are geometric with $a_1 = 6$ and $r = \frac{2}{3}$ . The third bounce uses $n - 1 = 2$ factors.

$a_3 = 6 \left(\frac{2}{3}\right)^{2} = 6 \cdot \frac{4}{9} = \frac{8}{3}$

Part B

After the initial $9$ -foot drop, every bounce rises and falls the same height, so the total is $9$ plus twice the infinite geometric sum of the rise heights, with $a_1 = 6$ and $r = \frac{2}{3}$ ( $|r| < 1$ ).

$9 + 2 \cdot \frac{6}{1 - \frac{2}{3}} = 9 + 2 \cdot 18 = 45$

Part C

The bounces never stop, but each is a fixed fraction of the last, so the rise heights shrink geometrically.

$|r| = \frac{2}{3} < 1 \quad\Longrightarrow\quad \text{the total settles rather than growing without bound}$

Were the bounces to return to the same height each time ( $|r| = 1$ ) or higher, the distance would grow without limit. The finite total is a direct consequence of $|r| < 1$ .

In one line

The bounce heights are geometric with $a_1 = 6$ feet and $r = \frac{2}{3}$ , so the third bounce rises to $\frac{8}{3}$ feet, and the total distance is $9 + 2 \cdot 18 = 45$ feet. Infinitely many bounces add to a finite distance because $|r| = \frac{2}{3} < 1$ makes the rise heights shrink toward zero, so the running total settles.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Identifies the bounce heights as geometric with $a_1 = 6$ , $r = \frac{2}{3}$ , and applies $n - 1 = 2$ factors. . Worth 2 points.

Reports the height with the unit, feet. . Worth 1 point.

Part B 4 points

Sets up the total as the drop plus twice the infinite geometric sum of the rise heights, checking $|r| < 1$ . . Worth 2 points.

Evaluates the infinite sum and combines it with the initial drop correctly. . Worth 1 point.

Reports the total distance with the unit, feet. . Worth 1 point.

Part C 3 points

Explains that the shrinking geometric rise heights let infinitely many bounces add to a finite total, tying it to $|r| < 1$ . . Worth 2 points. needs an explanation, not just an answer

Notes that a ratio with $|r| \ge 1$ would make the distance grow without bound. . Worth 1 point.
10. Two ways a sum can go on forever and still add up . 11 points. Question 10 of 10.

Two infinite series are given: a geometric series $\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{27} + \cdots$ , and a telescoping series $\displaystyle\sum_{k=1}^{\infty}\left(\frac{1}{k} - \frac{1}{k+1}\right)$ . This question sums each, then compares what makes each total exist.
1. Part A.
  Find the sum of the geometric series.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Find the sum of the telescoping series.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Each series adds infinitely many positive terms to a finite total, but the condition that guarantees this is stated differently for the two. State the condition for each, and explain why both come down to the same underlying requirement on the amount still left to add.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer

Part A

$\dfrac{1}{2}$ .

Part B

$1$ : the finite sums $1 - \dfrac{1}{n+1}$ approach $1$ as $n$ grows.

Part C

The geometric series needs $|r| < 1$ ; the telescope needs its surviving tail $b_{n+1}$ to shrink to zero. Both share one requirement on what is left to add: that leftover must shrink to zero, so the total settles. The individual terms shrinking is not enough, as the square-root telescope shows.

Worked solution

Part A

The ratio is $r = \frac{1}{3}$ , and $|r| < 1$ , so the infinite-sum formula applies.

$S = \frac{1/3}{1 - 1/3} = \frac{1/3}{2/3} = \frac{1}{2}$

Part B

With $b_k = \frac{1}{k}$ , the sum collapses to $b_1 - b_{n+1}$ , whose tail shrinks to zero.

$\sum_{k=1}^{n}\left(\frac{1}{k} - \frac{1}{k+1}\right) = 1 - \frac{1}{n+1} \;\longrightarrow\; 1$

Part C

For the geometric series the stated condition is $|r| < 1$ ; for the telescoping series it is that the surviving tail term shrinks to zero.

$\text{geometric: } |r| < 1 \;\Rightarrow\; a_1 r^{n} \to 0, \qquad \text{telescoping: } b_{n+1} \to 0$

Both say the same thing about what is left to add: the remaining contribution must shrink toward zero. When it does, the running total settles on a finite value; when it does not, the total grows or refuses to settle.

In one line

The geometric series sums to $\frac{1}{2}$ (since $|r| = \frac{1}{3} < 1$ ) and the telescoping series sums to $1$ (since $1 - \frac{1}{n+1} \to 1$ ). Their conditions, $|r| < 1$ and a vanishing surviving tail, are one requirement on what is left to add: that leftover must shrink to zero so the total settles. The individual terms shrinking is not enough (the square-root telescope's terms shrink yet its sum runs off).

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Checks $|r| < 1$ and applies the infinite geometric-sum formula. . Worth 2 points.

Reports the value as the settled total of the whole series. . Worth 1 point.

Part B 3 points

Collapses the telescoping sum to its correct closed form. . Worth 2 points.

States that the tail shrinks to zero, so the sum settles. . Worth 1 point.

Part C 5 points

States the correct condition for each series: $|r| < 1$ for the geometric, a vanishing surviving tail for the telescoping. . Worth 3 points. needs an explanation, not just an answer

Ties both conditions to the leftover amount still to be added shrinking to zero, distinct from the individual terms shrinking. . Worth 2 points. needs an explanation, not just an answer

Marking rubrics

Work every question first. Then mark your own paper against these, giving yourself the points for each element you actually wrote down.

1. A sequence that falls by a fixed step 9 points

Part A 3

Finds $d$ as a later term minus the earlier one, and applies $n - 1 = 11$ steps rather than $12$ . 2
Reports the computed value as the twelfth term itself, not its position. 1

Part B 3

Sets $a_n = -100$ and solves the resulting equation for $n$ . 2
Bases the membership verdict on whether $n$ is a positive whole number. 1

Part C 3

Shows the transition from $a_1$ to $a_2$ is one step, and generalizes that position $n$ takes $n-1$ steps, not $n$ . 3

2. Two terms, and how many ratios fit them 9 points

Part A 3

Sets up $a_4 = a_2 r^{2}$ with the correct exponent $4 - 2$ and isolates $r^{2}$ . 2
Reports both values of $r$ , not just the positive one. 1

Part B 3

Recovers $a_1$ from $a_2$ by dividing by each carried-forward ratio. 2
Pairs each first term with the ratio that produced it, keeping the two cases distinct. 1

Part C 3

Identifies the even distance as the source of the two solutions, tying it to the even power hiding the sign. 2
States that an odd distance would pin $r$ to a single value. 1

3. One series totalled two ways, and a stretch in the middle 10 points

Part A 3

Finds the last term and applies a correct sum formula with $n = 20$ . 2
Reports the result as the total of all twenty terms, not a single term. 1

Part B 4

Counts the terms in the range inclusively, using $n - m + 1$ rather than $n - m$ . 2
Finds the range's own first and last term and totals it correctly. 1
Reports the value as the total of the whole range, not a single term. 1

Part C 3

Explains that both endpoints are counted, so plain subtraction leaves one out, giving $n - m + 1$ . 2
States that the range is itself an arithmetic series, so the ordinary formula applies. 1

4. A telescoping evaluation, checked line by line 10 points

Part A 3

Names the first invalid line and clears the lines before it as sound. 1
Explains, from the gap-2 survivor rule, why two pieces survive at each end. 2

Part B 4

Combines both front survivors and both back survivors, not one of each. 2
Simplifies the front pieces and reports the full corrected closed form. 2

Part C 3

Ties the survivor count to how many brackets apart a term meets its cancelling partner, contrasting gap-1 with gap-2. 2
States the reasoning holds for any gap-2 telescope, not just this one. 1

5. A finite geometric total, and the case the formula cannot touch 9 points

Part A 3

Reads off $a_1$ and $r$ and applies the finite geometric-sum formula with $n = 5$ . 2
Reports the result as the total of all five terms, not one of them. 1

Part B 3

Recognizes $r = 1$ and totals with $S_n = n\,a_1$ rather than the ratio-based formula. 2
Reports the total as the term added to itself seven times. 1

Part C 3

Ties the breakdown to the denominator $r - 1$ being zero at $r = 1$ , making the formula undefined. 2
States the replacement rule $S_n = n\,a_1$ for a constant series. 1

6. Two savings plans, two kinds of growth 10 points

Part A 3

Classifies Avery's deposits as arithmetic and Blair's as geometric by the constant-difference versus constant-ratio test. 2
States the first term and the common difference or ratio for each plan. 1

Part B 4

Finds Avery's day-6 deposit with $n - 1 = 5$ steps of the common difference. 1
Finds Blair's day-6 deposit with $n - 1 = 5$ factors of the common ratio. 2
Reports both amounts with the unit, dollars. 1

Part C 3

Distinguishes the two by testing for a constant difference versus a constant ratio, dividing rather than subtracting for the geometric one. 2
Describes the long-run behavior of each: steady increase versus increase that multiplies without bound. 1

7. An infinite total, and the condition it depends on 9 points

Part A 3

Finds $r$ by dividing and checks $|r| < 1$ before applying the formula. 2
Reports the result as the settled total of the whole infinite series. 1

Part B 3

Reads off the first term and ratio from the place value of the repeating block. 1
Sums to a fraction and reduces it to lowest terms. 2

Part C 3

Ties the $|r| < 1$ requirement to the terms shrinking so the running total settles. 2
States that the formula still outputs a number for $|r| \ge 1$ , but it is not a genuine sum. 1

8. Two telescoping tails, and which one settles 9 points

Part A 3

Collapses the first sum to its correct closed form. 2
States that the tail $\frac{1}{n+4}$ shrinks to zero, so the form approaches $\frac{1}{4}$ . 1

Part B 3

Collapses the second sum to its correct closed form. 2
States that $\sqrt{n+1}$ grows without bound, so the form has no limiting value. 1

Part C 3

Identifies the shrinking of the surviving tail as the deciding feature, not the telescoping itself. 2
Applies it to rule the first sum in and the second out. 1

9. A bouncing ball and the distance it travels 10 points

Part A 3

Identifies the bounce heights as geometric with $a_1 = 6$ , $r = \frac{2}{3}$ , and applies $n - 1 = 2$ factors. 2
Reports the height with the unit, feet. 1

Part B 4

Sets up the total as the drop plus twice the infinite geometric sum of the rise heights, checking $|r| < 1$ . 2
Evaluates the infinite sum and combines it with the initial drop correctly. 1
Reports the total distance with the unit, feet. 1

Part C 3

Explains that the shrinking geometric rise heights let infinitely many bounces add to a finite total, tying it to $|r| < 1$ . 2
Notes that a ratio with $|r| \ge 1$ would make the distance grow without bound. 1

10. Two ways a sum can go on forever and still add up 11 points

Part A 3

Checks $|r| < 1$ and applies the infinite geometric-sum formula. 2
Reports the value as the settled total of the whole series. 1

Part B 3

Collapses the telescoping sum to its correct closed form. 2
States that the tail shrinks to zero, so the sum settles. 1

Part C 5

States the correct condition for each series: $|r| < 1$ for the geometric, a vanishing surviving tail for the telescoping. 3
Ties both conditions to the leftover amount still to be added shrinking to zero, distinct from the individual terms shrinking. 2

Multiple choice

Free response