Algebra 3 Quiz: Arithmetic Sequences And Series

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Arithmetic Sequences And SeriesQuestion 1 of 12

An arithmetic sequence is given by $a_n = 4n - 7$ . What is $S_{25}$ , the sum of the first 25 terms?

1125

1175

1075

2250

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Algebra 3 Quiz

Algebra 3 Quiz: Arithmetic Sequences And Series

Practice Arithmetic Sequences And Series in Algebra 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Arithmetic Sequences And Series, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

An arithmetic sequence is given by $a_n = 4n - 7$ . What is $S_{25}$ , the sum of the first 25 terms?

$1125$ (correct answer)
$1175$
$1075$
$2250$

Explanation: An arithmetic sequence has a constant difference between consecutive terms. Here,

a_n = 4n - 7

tells you directly that

a_1=4-7=-3

, and that the common difference is 4. For any arithmetic series, use the formula

S_n = \frac{n}{2}(a_1+a_n).

For

n=25

, compute the last term first:

a_{25}=4(25)-7=100-7=93. $$ $$ S_{25}=\frac{25}{2}(-3+93) =\frac{25}{2}(90) =25\cdot45=1125.

So 1125 is the correct sum. The wrong choices come from common slips.

2250

results from using

25(90)

instead of

\frac{25}{2}(90)

—forgetting the half.

1075

results from accidentally using the 24th term,

a_{24}=89

, as the last term:

\frac{25}{2}(-3+89)=1075

1175

results from going the other way and using the 26th term,

a_{26}=97

\frac{25}{2}(-3+97)=1175

. Both are off-by-one indexing errors. When you see a question asking for

S_{25}

, make sure the term index matches the number of terms: the last term included must be

a_{25}

, not

a_{24}

a_{26}

. A quick check of

a_1

a_{25}

, and the half in the formula will keep you accurate.

Question 2

In an arithmetic sequence, $a_1=5$ and $S_8=124$ . What is the common difference $d$ ?

$31$
$\frac{21}{8}$
$\frac{13}{4}$
$3$ (correct answer)

Explanation: An arithmetic sequence is linear: each term is the previous term plus

d

. When you know the first term, the number of terms, and the sum, the key formula is

S_n=\frac{n}{2}(2a_1+(n-1)d).

Plug in

n=8

a_1=5

, and

S_8=124

124=\frac{8}{2}(2(5)+7d)=4(10+7d)=40+28d.

84=28d

, giving

d=3

. The common difference is therefore

3

. The choice

31

is a trap: since

S_8=4(a_1+a_8)

, you get

a_1+a_8=31

, but that is the sum of the first and eighth terms, not the common difference. The choice

\frac{21}{8}

comes from using

a_8=a_1+8d

instead of

a_1+7d

, which gives

124=40+32d

. The choice

\frac{13}{4}

often comes from taking that

31

as if it were

a_8

, then computing

(31-5)/8=26/8=13/4

. That compounds two mistakes:

31

is not

a_8

, and there are only

7

gaps from

a_1

a_8

. Remember: with

n

terms, the last term uses

n-1

differences. Also,

\frac{2S_n}{n}

gives

a_1+a_n

, not

a_n

alone. Check your indexing before dividing.

Question 3

In an arithmetic sequence, the 3rd term is $19$ and the 8th term is $4$ . What is the sum of all positive terms?

$116$
$115$
$117$ (correct answer)
$112$

Explanation: Whenever you see an arithmetic sequence question, your first move is to find the common difference. Here, the 3rd term is

19

and the 8th term is

4

. Since there are

5

steps from term 3 to term 8, the difference is

\frac{4-19}{5}=-3

. Working backward, the first term is

19 - 2(-3)=25

, so the sequence is

25, 22, 19, \dots

. A term is positive as long as

25+(n-1)(-3)>0

, which gives

n<9\tfrac{1}{3}

, so the positive terms are terms 1 through 9. The 9th term is

25+8(-3)=1

. The sum of the first 9 terms is

\frac{9}{2}(25+1)=117

. The other choices come from stopping at the wrong place.

116

is the sum of the first 8 terms, so it leaves out the 9th term, which is still positive

1

115

is the sum of the first 10 terms, but the 10th term is

-2

, and negative terms should not be included.

112

is the sum of the first 7 terms, stopping before the positive 8th term

4

and 9th term

1

are added. For any arithmetic sequence sum question, first determine the common difference, then carefully identify exactly which terms satisfy the condition—here, being positive. Counting one term too few or too many is the most common trap.

Question 4

For an arithmetic sequence, $S_{10}=200$ and $S_{20}=800$ . What is $S_{30}$ ?

$1200$
$1400$
$1800$ (correct answer)
$1000$

Explanation: Whenever you see a question like this with

S_{10}}

S_{20}}

, and

S_{30}

, think in blocks of 10 terms. For an arithmetic sequence, sums over equal-sized blocks also form an arithmetic sequence. That's he key to this problem. Let the sums of he first, second, and third blocks of 10 terms be

B_1

B_2

B_3

. You know

B_1=S_{10}=200

. And

B_2=S_{20}-S_{10}=800-200=600

. Since block sums form an arithmetic sequence, he gap between them is constant:

600-200=400

. So the next block sum is

B_3=600+400=1000

. Therefore

S_{30}=S_{20}+B_3=800+1000=1800

. That is the correct result. The

1200

choice comes from scaling

800

30/20

, assuming the total is proportional to he number of terms. But the average per term is not constant in an arithmetic sequence; here it increases, so

1200

is too low. The

1400

choice comes from adding the change

S_{20}-S_{10}}=600

again to

S_{20}

. That treats the second block sum as if it were also he third block sum, but he third block is actually

1000

. The

1000

choice simply adds

S_{10}+S_{20}

, which has no valid meaning for partial sums. Study tip: for arithmetic sequences, always compare equal-sized blocks. The block sums arehemselves arithmetic, so use their constant difference — not he terms' common difference directly — to extend partial sums.

Question 5

The sum of the first $n$ terms of an arithmetic series is given by $S_n = 3n^2 - n$ . What is the common difference?

$3$
$6$ (correct answer)
$2$
$8$

Explanation: Whenever you see the sum of the first

n

terms of an arithmetic series written as a quadratic like

S_n = ̂3n^2 - n

, think about the general form. For an arithmetic series,

S_n = \frac n2\left[2a + (n-1)d\right] = \frac d2 n^2 + \left(a - \frac d2\right)n.

So the coefficient of

n^2

\frac d2

, not simply

d

. Here the coefficient is

3

, so

\frac d2 = ̂3

, which gives

d=6

. That is why the correct answer is

6

. The distractor

3

is just the coefficient of

n^2

; students often mistake it for the common difference, but it is only half of

d

. The choice

2

might come from confusing the coefficient of

n

with the difference, or from dividing

6

by something incorrectly;it has no direct role in the sum formula's leading coefficient. The choice

8

could result from an arithmetic slip like

3\cdot 2 + ̂2

, but there is no support for it in the comparison. For your study tip: on this exam, whenever an arithmetic series sum is given as

S_n = an^2 + bn

, the common difference is always

2a

. Remember the factor of

2

comes from the formula's

\frac d2 n^2

. Then check the linear coefficient if needed to find the first term, but the difference is determined entirely by the quadratic coefficient.

Question 6

In the arithmetic sequence $a_n = 3n + 2$ , what is the sum of the terms from $a_6$ through $a_{10}$ , inclusive?

$156$
$130$ (correct answer)
$185$
$55$

Explanation: Whenever you need the sum of consecutive terms in an arithmetic sequence, your first move is to identify the endpoints and the number of terms. For an inclusive range, the count is last index minus first index plus one. Here,

a_6 = 3(6)+2 = 20

and

a_{10}=3(10)+2=32

. From

a_6

through

a_{10}

inclusive there are

10-6+1=5

terms, and the average of the first and last terms is

\frac{20+32}{2}=26

. So the sum is

5\cdot 26=130

. The distractor

156

comes from an off-by-one error: if you mistakenly count 6 terms,

6\cdot 26=156

. Remember that

a_6, a_7, a_8, a_9, a_{10}

is only five terms. The choice

185

is the sum of the first ten terms,

a_1

through

a_{10}

, since

\frac{10(5+32)}{2}=185

; that includes

a_1

through

a_5

, which are not wanted. The choice

55

is the sum of the first five terms,

a_1

through

a_5

, because

\frac{5(5+17)}{2}=55

; it starts at the wrong endpoint. For inclusive sums, always calculate the term count as

n = \text{last} - \text{first} + 1

. If the range is small, listing the terms quickly (

20+23+26+29+32

) can also help you avoid endpoint errors.

Question 7

What is the sum of all integers from $-13$ through $47$ , inclusive?

$799$
$1020$
$1054$
$1037$ (correct answer)

Explanation: When you see "sum of all integers from

a

through

b

, inclusive," you're finding an arithmetic series. The key formula is:

\text{count} = b - a + 1, \quad \text{sum} = \text{count} \times \frac{a + b}{2}.

Here

a=-13

b=47

, so the count is

47 - (-13) + 1 = 61,

and the average of the first and last terms is

\frac{-13 + 47}{2} = \frac{34}{2} = 17.

Thus,

61 \times 17 = 1037.

That is the correct sum. Notice the pair structure: the 61 terms have middle value

17

, so your sum is

30

pairs of sum

34

plus the middle

17

, which also gives

30 \times 34 + 17 = 1037

. The choice

799

comes from using

47

as the number of terms:

47 \times 17 = 799

. But

47

is just the endpoint, not the count — the count includes the negative numbers and zero, so it must be

61

. The choice

1020

60 \times 17

. That happens if you compute the count as

47 - (-13) = 60

, forgetting to add

1

for "inclusive." You need that extra term, which is the middle

17

. The choice

1054

62 \times 17

. That trap comes from adding

1

twice — treating both endpoints as extra. But the difference

60

already measures gaps; for inclusive endpoints, add exactly one, giving

61

. Study takeaway: for any consecutive integer sum, always write the count first:

\text{last} - \text{first} + 1

. If your count times the average is around

1000

, you can sanity-check against

61 \times 17

Question 8

The 9th term of an arithmetic sequence is $-14$ , and the 17th term is $26$ . Which formula gives the nth term of this sequence?

$a_n=5n-59$ (correct answer)
$a_n=5n-54$
$a_n=5n-49$
$a_n=5n-64$

Explanation: Whenever a problem gives you two terms of an arithmetic sequence, your job is to find the common difference

d

and then the starting term

a_1

. Here the 17th and 9th terms are separated by

17-9=8

positions, and the value increases by

26-(-14)=40

. So

d=\frac{40}{8}=5

. Now use the 9th term to step back to

a_1

a_9=a_1+8(5)

, so

-14=a_1+40

, which gives

a_1=-54

. Thus the nth term is

a_n=a_1+(n-1)d=-54+5(n-1)=5n-59

. Plugging in

n=9

gives

-14

, and

n=17

gives

26

, so this formula matches the data. Each wrong choice has slope

5

but the wrong intercept. The formula

5n-54

comes from writing

a_n=-54+5n

instead of

-54+5(n-1)

; that forgets the crucial

(n-1)

shift. The formula

5n-49

is off by

+10

: substituting

n=9

gives

-4

, not

-14

. The formula

5n-64

is off by

-5

: substituting

n=9

gives

-19

. A common source of that last error is finding

a_1=26-17(5)=-59

by using

17

steps back from the 17th term, rather than

16

steps. That off-by-one mistake creates an intercept that is

5

too small. So when you solve this type of problem, always compute

d

first, then find

a_1

by stepping back with

(\text{term index}-1)d

. Finally, test your formula by plugging in one of the given term numbers—that catches intercept errors quickly.

Question 9

Five arithmetic means are inserted between $8$ and $44$ . What is the sum of the five inserted means?

$182$
$130$ (correct answer)
$156$
$174$

Explanation: Whenever you insert arithmetic means, think of them as terms of one arithmetic sequence: the endpoints are the first and last terms, and each inserted mean is a term in between. Here five means are inserted, so the sequence has

7

terms total. The common difference is

d=\frac{44-8}{5+1}=6.

So the five means are

14,20,26,32,38

, and their sum is

14+20+26+32+38=130.

Notice that the sum of all seven terms is

\frac{8+44}{2}\cdot 7=182

. That explains the trap behind 182: it is the total of everything, including both endpoints. The choice 156 would result from mistakenly inserting six means instead of five, since six means would have sum

6\cdot 26=156

. The choice 174 equals

14+20+26+32+38+44

, so it includes the last endpoint but still leaves out the first endpoint

8

. When you see "arithmetic means," set up the sequence immediately: first term is the lower endpoint, last term is the upper endpoint, and the number of terms is the number of means plus

2

. Then use

d=\frac{\text{last}-\text{first}}{\text{number of means}+1}

and list the means. This protects you from both endpoint and interval-counting mistakes.

Question 10

The sum of the first $n$ terms of the arithmetic sequence $5, 9, 13, \ldots$ is $860$ . What is $n$ ?

$18$
$19$
$21$
$20$ (correct answer)

Explanation: Whenever you see a question about the sum of an arithmetic sequence, start by naming the first term and common difference. Here,

a=5

and

d=4

, so the sum of the first

n

terms is

S_n=\frac{n}{2}\left(2(5)+(n-1)(4)\right)=\frac{n}{2}(4n+6)=2n^2+3n.

Set this equal to 860:

2n^2+3n=860 \Rightarrow 2n^2+3n-860=0.

Using the quadratic formula, the discriminant is

3^2-4(2)(-860)=6889=83^2

, so

n=\frac{-3\pm 83}{4}.

The positive solution is

n=\frac{80}{4}=20

, and the negative solution is discarded because

n

must be positive. To confirm, the 20th term is

5+19(4)=81

, and the average of the first and last terms is

\frac{5+81}{2}=43

. Multiplying by 20 terms gives

20\cdot 43=860

. Now consider the wrong answers. Choosing 18 gives a sum of

2(18)^2+3(18)=702

, which is too low. Choosing 19 gives

2(19)^2+3(19)=779

, still short of 860. Choosing 21 gives

2(21)^2+3(21)=945

, which overshoots the target. These errors often come from forgetting to use the

n-1

term inside the parentheses, or from solving the quadratic incorrectly. On exam day, remember: for arithmetic sums, identify

a

and

d

, write the formula, and set it equal to the given sum. Then solve the quadratic and reject any negative or non-integer root.

Question 11

An auditorium has 20 rows of seats. Row 1 contains 20 seats, row 9 contains 44 seats, and the number of seats in consecutive rows increases by a constant amount. What is the total number of seats in the auditorium?

$970$ (correct answer)
$1000$
$1940$
$2000$

Explanation: Whenever you see consecutive quantities increasing by a constant amount, you are working with an arithmetic sequence. Here the row counts are the terms, so find the common difference first. Row 1 has 20 seats and row 9 has 44 seats; from row 1 to row 9 is 8 steps, so each step is

(44-20)/8 = 3

seats per row. Thus row 20 has

20 + 19(3) = 77

seats. Since an arithmetic sequence sum depends only on the first and last terms, the total is

\frac{20}{2}(20+77) = 10(97) = 970

. The choice

1000

comes from a common trap: using 20 as the number of increases, so row 20 becomes

20 + 20(3) = 80

, then averaging 20 and 80. But there are only 19 increases between row 1 and row 20. The choice

1940

uses the correct 77 but forgets to divide by 2:

20(20+77)

gives twice the sum. The choice

2000

combines both mistakes: it uses row 20 = 80 and also fails to divide by 2, giving

20(20+80)

. On exam day, write out the pattern and check the number of intervals: from row 1 to row

n

n-1

increases. Then apply the sum formula, and don't forget the factor

\frac{n}{2}

Question 12

The second term of an arithmetic sequence is $9$ , and the fifth term is $21$ . What is the sum of the first 15 terms?

$555$
$525$
$495$ (correct answer)
$465$

Explanation: An arithmetic sequence is built by adding the same difference

d

each time. You are told the second term is

9

and the fifth term is

21

, so between those two terms there are

5-2=3

steps. That means

21-9=12

is equal to

3d

, so

d=4

. Work backward to the first term: since the second term is

9

, the first term is

9-4=5

. Now use the sum formula for the first 15 terms:

S_{15}=\frac{15}{2}\left[2(5)+(15-1)(4)\right] =\frac{15}{2}(10+56) =\frac{15}{2}(66)=495.

495

is the correct sum. The choice

555

comes from treating the second term,

9

, as if it were the first term and then summing 15 terms with that starting value:

\frac{15}{2}(18+14\cdot4)=555

. The choice

525

incorrectly combines the second term with the correct 15th term, using an average of

\frac{9+61}{2}

, which mixes up which term is the true starting point. The choice

465

would require the first term to be

3

, but then the second term would be

7

, not

9

; this is an off-by-one indexing error. The key is always to identify

a_1

and

d

before summing. Count the intervals between known terms carefully, then use

S_n=\frac n2(a_1+a_n)