Sequences and Series: Chapter Test

20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.

Multiple choice

20 questions, 100 points in total, 5 points each. Answer in any order and change your mind as often as you like. When you submit, your answers lock and every question shows its worked solution.

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Question 1 of 20

1

A sequence begins $8,\ 13,\ 18,\ 23,\ \ldots$ and each term after the first adds the same amount. What is $a_{11}$ ?

Answer choices for question 1 A $63$ B $58$ C $50$ D $53$
2

The sequence $5,\ 10,\ 20,\ 40,\ \ldots$ keeps multiplying by the same number. What is its sixth term?

Answer choices for question 2 A $320$ B $32$ C $160$ D $80$
3

Find $3 + 7 + 11 + \cdots + 83$ .

Answer choices for question 3 A $903$ B $1806$ C $860$ D $945$
4

Find the sum $2 + 6 + 18 + 54 + 162$ .

Answer choices for question 4 A $410$ B $728$ C $80$ D $242$
5

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Given that $\dfrac{1}{k(k+1)} = \dfrac{1}{k} - \dfrac{1}{k+1}$ for every positive integer $k$ , evaluate $\displaystyle\sum_{k=1}^{14} \dfrac{1}{k(k+1)}$ .

Answer choices for question 5 A $\dfrac{14}{15}$ B $\dfrac{1}{15}$ C $-\dfrac{14}{15}$ D $\dfrac{15}{16}$
6

In an arithmetic sequence, $a_3 = 20$ and $a_7 = 8$ . What is $a_1$ ?

Answer choices for question 6 A $23$ B $26$ C $14$ D $29$
7

A geometric sequence has $a_1 = 4$ and $a_5 = 324$ . Which statement about its common ratio $r$ is correct?

Answer choices for question 7 A $r = 3$ is the only possibility. B $r = 81$ . C $r$ cannot be found from only two terms. D $r$ must be $3$ or $-3$ .
8

Find the sum of the infinite series $18 + 6 + 2 + \dfrac{2}{3} + \cdots$ .

Answer choices for question 8 A $\dfrac{27}{2}$ B $54$ C $27$ D $\dfrac{80}{3}$
9

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Given $\dfrac{1}{k(k+2)} = \dfrac{1}{2}\left(\dfrac{1}{k} - \dfrac{1}{k+2}\right)$ , evaluate $\displaystyle\sum_{k=1}^{\infty} \dfrac{1}{k(k+2)}$ .

Answer choices for question 9 A $\dfrac{3}{4}$ B $\dfrac{3}{2}$ C $\dfrac{1}{2}$ D $1$
10

One of these lists is arithmetic and the other is geometric: $\;3,\ 7,\ 11,\ 15,\ \ldots\;$ and $\;3,\ 6,\ 12,\ 24,\ \ldots$ . Taking the geometric one, what is its seventh term?

Answer choices for question 10 A $27$ B $192$ C $384$ D $96$
11

An arithmetic sequence has $a_1 = 5$ and $d = 4$ . What is the sum of its $10$ th through $20$ th terms, inclusive?

Answer choices for question 11 A $610$ B $732$ C $671$ D $715$
12

Exactly one of these infinite series has a finite sum. Which one?

Answer choices for question 12 A $2 + 4 + 8 + 16 + \cdots$ B $7 + 7 + 7 + 7 + \cdots$ C $5 - 5 + 5 - 5 + \cdots$ D $9 + 3 + 1 + \dfrac{1}{3} + \cdots$
13

Advanced. This question goes beyond core Algebra I. It is not required by the course.

The sum $\displaystyle\sum_{k=1}^{n}\left(\dfrac{1}{k+1} - \dfrac{1}{k+2}\right)$ collapses to which closed form?

Answer choices for question 13 A $\dfrac{1}{2} - \dfrac{1}{n+2}$ B $1 - \dfrac{1}{n+2}$ C $\dfrac{1}{n+2} - \dfrac{1}{2}$ D $\dfrac{1}{2} - \dfrac{1}{n+1}$
14

Write the repeating decimal $0.\overline{18}$ as a fraction in lowest terms.

Answer choices for question 14 A $\dfrac{9}{50}$ B $\dfrac{2}{11}$ C $\dfrac{1}{5}$ D $\dfrac{18}{101}$
15

A geometric sequence has $a_1 = 2$ and $r = 3$ . Which of these is a term of the sequence?

Answer choices for question 15 A $243$ B $150$ C $30$ D $162$
16

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Two infinite sums are built the same way, by adding shrinking pieces: $\displaystyle\sum_{k=1}^{\infty}\left(\sqrt{k+1} - \sqrt{k}\right)$ and $\displaystyle\sum_{k=1}^{\infty}\left(\dfrac{1}{k} - \dfrac{1}{k+1}\right)$ . Which one has a finite value?

Answer choices for question 16 A Only the first. B Only the second. C Both, since both telescope to two survivors. D Neither.
17

To add $1 + 2 + 3 + \cdots + 100$ , one pairs the first term with the last, the second with the second-to-last, and so on; each of the $50$ pairs sums to $101$ , giving $5050$ . Using the same pairing idea, what is $1 + 2 + 3 + \cdots + 99$ ?

Answer choices for question 17 A $9900$ B $4900$ C $4950$ D $5000$
18

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Evaluate $\displaystyle\sum_{k=1}^{48} \dfrac{1}{\sqrt{k} + \sqrt{k+1}}$ .

Answer choices for question 18 A $7$ B $\sqrt{49} - \sqrt{48}$ C $6$ D $-6$
19

Find the sum of the infinite series $6 - 2 + \dfrac{2}{3} - \dfrac{2}{9} + \cdots$ .

Answer choices for question 19 A $\dfrac{9}{2}$ B $9$ C $\dfrac{40}{9}$ D $18$
20

The multiples of $7$ strictly between $100$ and $400$ are added together. What is their sum?

Answer choices for question 20 A $21672$ B $10584$ C $10750$ D $10836$

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Where you stand

Work with an arithmetic sequence: find a term with $a_n = a_1 + (n-1)d$ counting $n-1$ steps, recover $d$ and $a_1$ from given terms, and decide whether a value is a term. Review: Arithmetic Sequences
Sum an arithmetic series with $\frac{n(a_1+a_n)}{2}$ or $\frac{n}{2}(2a_1+(n-1)d)$ , and total an inclusive range of $n-m+1$ terms. Review: Arithmetic Series
Work with a geometric sequence: find the common ratio and a term with $a_n = a_1 r^{n-1}$ , recover $r$ from two terms with its sign ambiguity at even distance, and decide membership. Review: Geometric Sequences
Sum a finite geometric series with $\frac{a_1(r^n-1)}{r-1}$ (and $n a_1$ at $r=1$ ), and evaluate an infinite geometric series with $\frac{a_1}{1-r}$ exactly when $\lvert r\rvert < 1$ , including a repeating decimal. Review: Geometric Series
Collapse a telescoping sum to $b_1 - b_{n+1}$ , apply the gap-2 correcting factor with two survivors at each end, and decide whether an infinite telescope settles by whether its surviving tail shrinks to zero. Review: Telescoping Sums
Decide which family or tool a list or series calls for and whether an infinite process has a finite value, combining ideas across the chapter: classifying a list, reading a repeating decimal as a geometric series, and comparing a telescoping tail with a geometric one. Review: Arithmetic Sequences , Arithmetic Series , Geometric Sequences , Geometric Series , Telescoping Sums

Core practice

10 problems from across the chapter. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 The paired labels

An arithmetic sequence satisfies $a_2+a_6=28$ and $a_4+a_7=35$ . Find its first term and common difference.
1. Hint 1
  Terms equally far from the fourth position have average $a_4$ .
2. Hint 2
  Use the second condition to find $a_7$ .
3. Hint 3
  Three equal steps connect the fourth and seventh positions.
Answer

$a_1=7$ and $d=\frac73$ .

Full solution

The first condition gives $2a_4=28$ , so $a_4=14$ .
The second then gives $a_7=21$ .
There are three steps between them, so
$3d=21-14.$
Hence $d=7/3$ , and
$a_1=14-3\cdot\frac73.$
Thus $a_1=7$ .
Checking gives $a_2=28/3$ and $a_6=56/3$ , which total $28$ , while
$a_4+a_7=14+21=35.$

Answer

$a_1=7$ and $d=\frac73$ .

Key idea

Sums of equally spaced arithmetic terms can reveal a middle term before the common difference is recovered.
Problem 1: Mark complete
Problem 2 The third term

A geometric sequence with real terms starts with $a_1=2$ and satisfies $a_3=a_2+12$ . Find every possible common ratio.
1. Hint 1
  Express both later terms in terms of the common ratio.
2. Hint 2
  The condition produces a quadratic equation.
3. Hint 3
  Check each real solution in the original relation.
Answer

$r=3$ or $r=-2$ .

Full solution

The later terms are $a_2=2r$ and $a_3=2r^2$ .
Therefore
$2r^2=2r+12.$
Dividing by $2$ and rearranging gives
$r^2-r-6=0.$
This is $(r-3)(r+2)=0$ , so $r=3$ or $r=-2$ .
They produce the first three terms $2,6,18$ and $2,-4,8$ , respectively, and both satisfy
$a_3=a_2+12.$

Answer

$r=3$ or $r=-2$ .

Key idea

A relation among geometric terms can become an equation for the common ratio.
Problem 2: Mark complete
Problem 3 The remaining total

An infinite geometric series has total $24$ , and the sum of all its terms after the first is $16$ . Find its first term and its common ratio, and the sum of all its terms after the second.
1. Hint 1
  Everything after the first term is itself an infinite geometric series.
2. Hint 2
  The two given totals differ by exactly one term.
3. Hint 3
  Dropping one more term scales the remaining total by another factor of the ratio.
Answer

$a_1=8$ , $r=\frac23$ , and the terms after the second total $\frac{32}3$ .

Full solution

The whole series is its first term plus everything after it, so
$a_1=24-16.$
Thus $a_1=8$ .
Everything after the first term is the original series with every term multiplied by $r$ , so that part totals $24r$ .
Setting $24r=16$ gives $r=2/3$ , whose magnitude is less than $1$ , so the stated total is a genuine value.
Checking,
$8/(1-2/3)=24.$
Everything after the second term is the original series with every term multiplied by $r^2$ , so it totals
$24\cdot\frac49=\frac{32}3.$
As a check, $a_2=16/3$ , and $24-8-16/3$ is also $32/3$ .

Answer

$a_1=8$ , $r=\frac23$ , and the terms after the second total $\frac{32}3$ .

Key idea

The part of an infinite geometric series past a given position is the whole series scaled by a power of the ratio.
Problem 3: Mark complete
Problem 4 The adjusted report

Advanced. This question goes beyond core Algebra I. It is not required by the course.

For an integer $n\ge2$ , a report gives
$R_n=2\sum_{k=2}^{n}\frac{1}{k(k+2)}.$
It then adds $1/(n+1)$ and $1/(n+2)$ to $R_n$ . Find the final value, in simplest form.
1. Hint 1
  Combine fractions to rewrite each summand as a difference.
2. Hint 2
  The factor $2$ cancels the correcting factor in that difference.
3. Hint 3
  Track the two boundary fractions at each end.
Answer

$\frac56$ .

Full solution

For positive $k$ , all denominators are nonzero and
$\frac{2}{k(k+2)}=\frac1k-\frac1{k+2}.$
Write
$B=1/(n+1)+1/(n+2).$
Because the sum starts at $k=2$ , the gap of two leaves the positive front values $1/2$ and $1/3$ and the two negative end values making up $B$ .
Therefore
$R_n=\frac56-B.$
The report then adds $B$ , giving $5/6$ .
For $n=2$ the report is $R_2=1/4$ and $B=7/12$ , and $1/4+7/12=5/6$ , which also checks the short case.

Answer

$\frac56$ .

Key idea

Adding back all surviving end fractions can make a telescoping report independent of its stopping index.
Problem 4: Mark complete
Problem 5 The package range

Package weights form an arithmetic sequence. Package $2$ weighs $7$ grams and package $5$ weighs $16$ grams. The packages from $4$ through $N$ , inclusive, have total weight $95$ grams. Find $N$ .
1. Hint 1
  Recover the common difference from the known weights.
2. Hint 2
  Let the number of included packages be $N-4+1$ .
3. Hint 3
  Use the first included weight and the arithmetic total to find the count.
Answer

$N=8$ .

Full solution

Three steps increase the weight by $16-7=9$ grams, so $d=3$ grams.
Package $4$ weighs $13$ grams.
Let $m=N-3$ be the positive included count.
The total condition gives
$\frac{m(26+3(m-1))}{2}=95.$
Rearranging gives
$3m^2+23m-190=0.$
Its discriminant is
$23^2+4\cdot3\cdot190=2809.$
Since $2809=53^2$ , the quadratic formula gives
$m=\frac{-23+53}{6}.$
So $m=5$ , the other root being negative and therefore impossible for a count, and $N=8$ .
The included weights $13,16,19,22,25$ total $95$ grams, confirming both endpoints.

Answer

$N=8$ .

Key idea

An inclusive arithmetic total may determine the last position after the common difference has been recovered.
Problem 5: Mark complete
Problem 6 The separated terms

A geometric sequence with real terms has $a_3=1$ and $a_5=\frac1{16}$ . For each possible common ratio, decide whether the sum of all its terms, starting at $a_1$ , exists, and give that sum where it does.
1. Hint 1
  The two known positions are two ratio steps apart.
2. Hint 2
  Keep both real signs when recovering the ratio.
3. Hint 3
  Find the first term for each ratio and check convergence before summing.
Answer

The ratios are $r=\frac14$ and $r=-\frac14$ , and the sum exists for both: it is $\frac{64}3$ when $r=\frac14$ and $\frac{64}5$ when $r=-\frac14$ .

Full solution

The two given positions are two steps apart, so $a_5=a_3r^2$ gives $r^2=1/16$ .
Thus $r=1/4$ or $r=-1/4$ .
Each has magnitude less than $1$ , so the sum of all the terms exists in both cases.
Either sign gives the same first term, since $a_1=a_3/r^2$ depends on $r^2$ alone, so $a_1=16$ .
With $r=1/4$ ,
$S=\frac{16}{1-1/4}.$
This gives $S=64/3$ .
With $r=-1/4$ ,
$S=\frac{16}{1+1/4}.$
This gives $S=64/5$ .
Both ratios give $a_3=1$ and $a_5=1/16$ , so neither can be discarded.

Answer

The ratios are $r=\frac14$ and $r=-\frac14$ , and the sum exists for both: it is $\frac{64}3$ when $r=\frac14$ and $\frac{64}5$ when $r=-\frac14$ .

Key idea

An even gap between geometric terms can leave two ratios that produce different infinite totals.
Problem 6: Mark complete
Problem 7 The difference rule

A sequence is defined by $a_n=5^{n+1}-5^n$ for positive integers $n$ . A student calls it arithmetic because its formula subtracts two expressions. Decide whether the sequence is arithmetic, geometric, both, or neither, and determine whether $2500$ is one of its terms.
1. Hint 1
  Simplify the formula by taking out the shared power.
2. Hint 2
  Compare neighboring terms after simplifying.
3. Hint 3
  Solve the term equation and check that the position is a positive integer.
Answer

Geometric only; $2500=a_4$ .

Full solution

Taking out the common power gives
$a_n=4\cdot5^n.$
Each term is five times the preceding term, so the sequence is geometric.
Its first three terms are $20,100,500$ ; their differences $80$ and $400$ are unequal, so it is not arithmetic.
To test the requested value, solve
$4\cdot5^n=2500.$
This gives $5^n=625$ , and $625=5^4$ , hence $n=4$ .
Substitution gives
$5^5-5^4=3125-625=2500.$

Answer

Geometric only; $2500=a_4$ .

Key idea

A sequence family is determined by relationships among its terms, not by the operation visible in its original formula.
Problem 7: Mark complete
Problem 8 The two records

One record is the number $X=0.\overline{24}$ . Another is the three-term total $Y=\frac18+\frac1{16}+\frac1{32}$ . Which record is larger, and by exactly how much?
1. Hint 1
  An exact comparison needs both records written as fractions before anything is subtracted.
2. Hint 2
  Identify the first contribution and ratio for each record.
3. Hint 3
  Compute the exact totals, then compare them using a common denominator.
Answer

$X$ is larger by $\frac{25}{1056}$ .

Full solution

For $X$ , each two-digit block contributes $24/100$ times a power of $1/100$ .
This ratio has magnitude less than $1$ , so
$X=\frac{24/100}{1-1/100}.$
Thus $X=8/33$ .
The record $Y$ has $n=3$ terms, first term $1/8$ and ratio $1/2$ , and it is the ratio that carries the exponent $3$ :
$Y=\frac{\tfrac18\left(1-\left(\tfrac12\right)^{3}\right)}{1-\tfrac12}.$
Thus $Y=7/32$ .
As a direct check, the three terms have numerators $4,2,1$ over $32$ .
The common denominator is $1056$ , so
$X-Y=\frac{256-231}{1056}.$
This is $25/1056>0$ , so $X$ is larger by exactly that amount.

Answer

$X$ is larger by $\frac{25}{1056}$ .

Key idea

Comparing totals requires checking how many terms each pattern includes before computing exact fractions.
Problem 8: Mark complete
Problem 9 The matching totals

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Find the positive constant $c$ for which these two infinite totals are equal:
$A=\sum_{k=1}^{\infty}\frac{c}{(2k-1)(2k+1)}$
and
$B=\sum_{k=1}^{\infty}\frac5{3^k}.$
Justify that both totals exist for your value.
1. Hint 1
  For the first total, look for cancellation between the reciprocals of consecutive odd numbers.
2. Hint 2
  For the second, identify its first term and constant ratio.
3. Hint 3
  Equate the two resulting totals.
Answer

$c=5$ ; both totals equal $\frac52$ .

Full solution

The two factors $2k-1$ and $2k+1$ differ by $2$ , so combining $1/(2k-1)-1/(2k+1)$ over a common denominator gives twice the summand of $A$ at $c=1$ .
The split therefore carries a correcting factor $1/2$ , making each summand of $A$ equal to $\frac c2\left(\frac1{2k-1}-\frac1{2k+1}\right)$ , with no denominator zero for $k\ge1$ .
Consecutive brackets cancel, leaving the front value $1$ and the end value $1/(2n+1)$ , so the finite partial total is
$A_n=\frac c2\left(1-\frac1{2n+1}\right).$
For any fixed positive $c$ , the final fraction shrinks to zero, so $A=c/2$ .
The second series has first term $5/3$ and ratio $1/3$ , which has magnitude less than $1$ .
Therefore
$B=\frac{5/3}{1-1/3}.$
Thus $B=5/2$ .
Equality requires $c/2=5/2$ , so $c=5$ , a positive value.
With this value both partial totals approach $5/2$ , as required.

Answer

$c=5$ ; both totals equal $\frac52$ .

Key idea

Totals produced by different series structures can be matched after each convergence condition is checked.
Problem 9: Mark complete
Problem 10 The closer display

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Two displays show
$A_n=\sum_{k=1}^{n}\frac{1}{k(k+1)}$
and
$B_n=1-2^{-n}$
for positive integers $n$ . Kai says both displays approach $1$ , and the second is closer to $1$ at $n=3$ . Is each part of the claim correct? Give the exact distances from $1$ at $n=3$ and justify your conclusions.
1. Hint 1
  Find the boundary expression for $A_n$ .
2. Hint 2
  Subtract each display from $1$ .
3. Hint 3
  Compare the two positive missing amounts at $n=3$ and describe their long-term behavior.
Answer

Both claims are correct; at $n=3$ the missing amounts are $\frac14$ and $\frac18$ , respectively.

Full solution

Splitting $1/[k(k+1)]$ as $1/k-1/(k+1)$ cancels the interior fractions, leaving
$A_n=1-\frac1{n+1}.$
Thus $A_n$ is missing $1/(n+1)$ from $1$ , while $B_n$ is missing $2^{-n}$ .
Each positive missing amount tends to zero, so both displays approach $1$ .
At $n=3$ , the amounts are
$1-A_3=\frac14$
and
$1-B_3=\frac18.$
Since $1/8<1/4$ , the second display is closer at that index.

Answer

Both claims are correct; at $n=3$ the missing amounts are $\frac14$ and $\frac18$ , respectively.

Key idea

Comparing a telescoping boundary with a geometric tail reveals both the limiting totals and the accuracy of particular partial totals.
Problem 10: Mark complete

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