Math 1 Quiz: Arithmetic Sequences
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Arithmetic SequencesQuestion 1 of 9

Consider the sequence an=5n−2a_n = 5n - 2. Which statement about this sequence is NOT true?

The sequence is arithmetic with common difference 55
The first term of the sequence is 33
The 1010th term minus the 77th term equals 1515
The sequence contains both positive and negative terms
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Math 1 Quiz

Math 1 Quiz: Arithmetic Sequences

Practice Arithmetic Sequences in Math 1 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, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 1.

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

Consider the sequence an=5n−2a_n = 5n - 2. Which statement about this sequence is NOT true?

  1. The sequence is arithmetic with common difference 55
  2. The first term of the sequence is 33
  3. The 1010th term minus the 77th term equals 1515
  4. The sequence contains both positive and negative terms (correct answer)
Explanation: For an=5n−2a_n = 5n - 2: Choice A is true since an+1−an=5(n+1)−2−(5n−2)=5a_{n+1} - a_n = 5(n+1) - 2 - (5n - 2) = 5. Choice B is true since a1=5(1)−2=3a_1 = 5(1) - 2 = 3. Choice C is true since a10−a7=(48)−(33)=15a_{10} - a_7 = (48) - (33) = 15. Choice D is false because for n≥1n \geq 1, we have an=5n−2≥5(1)−2=3>0a_n = 5n - 2 \geq 5(1) - 2 = 3 > 0, so all terms are positive. Students might think negative terms exist by considering n≤0n \leq 0, but sequence indices start at n=1n = 1.

Question 2

Two arithmetic sequences have the same first term. The first sequence has a common difference of 33 and its 88th term is 2929. The second sequence has a common difference of −2-2 and its kkth term equals the 1212th term of the first sequence. What is the value of kk?

  1. 18
  2. 19 (correct answer)
  3. 17
  4. 16
Explanation: For the first sequence: a8=a1+7(3)=29a_8 = a_1 + 7(3) = 29, so a1=8a_1 = 8. The 12th term is a12=8+11(3)=41a_{12} = 8 + 11(3) = 41. For the second sequence with a1=8a_1 = 8 and d=−2d = -2: ak=8+(k−1)(−2)=41a_k = 8 + (k-1)(-2) = 41. Solving: 8−2k+2=418 - 2k + 2 = 41, so 10−2k=4110 - 2k = 41, giving k=19k = 19. Choice A uses k−1=17k-1 = 17 without solving completely. Choice C results from sign error in the common difference. Choice D comes from using the wrong term from the first sequence.

Question 3

Consider the arithmetic sequence 2,7,12,17,22,...2, 7, 12, 17, 22, ... A student claims that 157 is a term in this sequence. To verify this claim, the student should check whether which of the following equations has an integer solution?

  1. 157=2+5n157 = 2 + 5n where n≥0n \geq 0
  2. 157=2+5(n−1)157 = 2 + 5(n-1) where n≥1n \geq 1 (correct answer)
  3. 157=5n−3157 = 5n - 3 where n≥1n \geq 1
  4. 157=7+5(n−2)157 = 7 + 5(n-2) where n≥2n \geq 2
Explanation: The arithmetic sequence has first term a1=2a_1 = 2 and common difference d=5d = 5. The general term is an=2+(n−1)⋅5a_n = 2 + (n-1) \cdot 5 where n≥1n \geq 1 represents the position in the sequence. To check if 157 is a term, we need 157=2+5(n−1)157 = 2 + 5(n-1) for some positive integer nn. Solving: 155=5(n−1)155 = 5(n-1), so n−1=31n-1 = 31 and n=32n = 32. Since this gives an integer solution, 157 is indeed the 32nd term. Choice A uses n≥0n \geq 0 which would make the first term correspond to n=0n = 0, shifting the indexing incorrectly. Choice C represents a different sequence entirely. Choice D incorrectly uses 7 as the reference point and adjusts the indexing inappropriately.

Question 4

A theater has rows of seats arranged so that each row has 4 more seats than the row in front of it. If the first row has ss seats and there are nn rows total, which expression represents the number of seats in the row that is exactly in the middle when nn is odd?

  1. s+4(n−12)s + 4\left(\frac{n-1}{2}\right) (correct answer)
  2. s+4(n+12)s + 4\left(\frac{n+1}{2}\right)
  3. s+2(n−1)s + 2(n-1)
  4. s+2ns + 2n
Explanation: Since each row has 4 more seats than the previous row, this forms an arithmetic sequence with first term ss and common difference d=4d = 4. The general term is ak=s+(k−1)⋅4a_k = s + (k-1) \cdot 4 where kk is the row number. When nn is odd, the middle row is row number n+12\frac{n+1}{2}. So the number of seats in the middle row is an+12=s+(n+12−1)⋅4=s+(n+1−22)⋅4=s+4(n−12)a_{\frac{n+1}{2}} = s + \left(\frac{n+1}{2} - 1\right) \cdot 4 = s + \left(\frac{n+1-2}{2}\right) \cdot 4 = s + 4\left(\frac{n-1}{2}\right). Choice B incorrectly uses n+12\frac{n+1}{2} without subtracting 1 in the arithmetic sequence formula. Choice C uses the wrong coefficient (2 instead of 4). Choice D also uses the wrong coefficient and doesn't account for the position properly.

Question 5

Two arithmetic sequences have the same first term of 8. Sequence A has a common difference of 3, and sequence B has a common difference of 5. After how many terms will sequence B first exceed sequence A by more than 50?

  1. 25
  2. 26
  3. 27 (correct answer)
  4. 28
Explanation: Sequence A: an(A)=8+(n−1)⋅3=3n+5a_n^{(A)} = 8 + (n-1) \cdot 3 = 3n + 5. Sequence B: an(B)=8+(n−1)⋅5=5n+3a_n^{(B)} = 8 + (n-1) \cdot 5 = 5n + 3. We want the smallest nn such that an(B)−an(A)>50a_n^{(B)} - a_n^{(A)} > 50. This gives us (5n+3)−(3n+5)>50(5n + 3) - (3n + 5) > 50, which simplifies to 2n−2>502n - 2 > 50, so 2n>522n > 52, and n>26n > 26. Therefore, n=27n = 27 is the first integer value where sequence B exceeds sequence A by more than 50. Let's verify: a27(B)=5(27)+3=138a_{27}^{(B)} = 5(27) + 3 = 138 and a27(A)=3(27)+5=86a_{27}^{(A)} = 3(27) + 5 = 86. The difference is 138−86=52>50138 - 86 = 52 > 50 ✓. For n=26n = 26: a26(B)=133a_{26}^{(B)} = 133 and a26(A)=83a_{26}^{(A)} = 83, difference = 50, which is not greater than 50.

Question 6

An arithmetic sequence has first term aa and common difference d>0d > 0. If exactly three terms of this sequence are two-digit numbers (from 10 to 99), which of the following relationships must be true?

  1. 90≤a≤9990 \leq a \leq 99 and d≥30d \geq 30
  2. 10≤a≤9710 \leq a \leq 97 and d≤45d \leq 45
  3. a<10a < 10 and 90<2d≤9990 < 2d \leq 99
  4. a≤97a \leq 97 and d>45d > 45 (correct answer)
Explanation: For exactly three terms to be two-digit numbers, we need the sequence to enter the two-digit range and then exit it after exactly three terms. Let's say the three two-digit terms are an,an+1,an+2a_n, a_{n+1}, a_{n+2} where 10≤an,an+1,an+2≤9910 \leq a_n, a_{n+1}, a_{n+2} \leq 99 but an−1<10a_{n-1} < 10 (if it exists) and an+3>99a_{n+3} > 99. This means: an+2≤99<an+3=an+2+da_{n+2} \leq 99 < a_{n+3} = a_{n+2} + d. So an+2≤99a_{n+2} \leq 99 and an+2+d>99a_{n+2} + d > 99, which gives us 99−d<an+2≤9999 - d < a_{n+2} \leq 99. Since an+2=an+2da_{n+2} = a_n + 2d and an≥10a_n \geq 10, we have an+2d≤99a_n + 2d \leq 99, so an≤99−2da_n \leq 99 - 2d. Also, an+2d>99−da_n + 2d > 99 - d, so an>99−3da_n > 99 - 3d. For the sequence to have exactly three two-digit terms, we need an≥10a_n \geq 10 but an−1=an−d<10a_{n-1} = a_n - d < 10, giving us 10−d<an<10+d10 - d < a_n < 10 + d (approximately). The key insight is that dd must be large enough that the sequence jumps from single digits to three-digit numbers in exactly three steps. If the first two-digit term is close to 10 and we need exactly three terms before exceeding 99, then 10+2d≤9910 + 2d \leq 99 and 10+3d>9910 + 3d > 99. This gives d≤44.5d \leq 44.5 and d>29.67d > 29.67. But this contradicts choice D. Let me reconsider: if aa itself could be two-digit and we want exactly three terms total to be two-digit, then a≤99a \leq 99, a+d≤99a + d \leq 99, a+2d≤99a + 2d \leq 99, but a+3d>99a + 3d > 99. The last condition gives d>99−a3d > \frac{99-a}{3}. For this to be satisfied with d>45d > 45, we need 45>99−a345 > \frac{99-a}{3}, so 135>99−a135 > 99 - a, giving a>−36a > -36, which is always true for reasonable values. The condition a≤97a \leq 97 ensures that when d>45d > 45, we can have exactly three terms in the two-digit range.

Question 7

An arithmetic sequence has the property that when each term is squared, the resulting sequence is 4,25,64,121,196,…4, 25, 64, 121, 196, \ldots. What is the common difference of the original arithmetic sequence?

  1. 3
  2. -3
  3. 3 or -3 (correct answer)
  4. Cannot be determined
Explanation: The squared sequence is 4,25,64,121,196,…=22,52,82,112,142,…4, 25, 64, 121, 196, \ldots = 2^2, 5^2, 8^2, 11^2, 14^2, \ldots. This means the original sequence could be 2,5,8,11,14,…2, 5, 8, 11, 14, \ldots (with common difference 33) or −2,−5,−8,−11,−14,…-2, -5, -8, -11, -14, \ldots (with common difference −3-3). Both sequences are arithmetic and produce the same squared sequence. Choice A only considers positive terms. Choice B only considers negative terms. Choice D incorrectly assumes the information is insufficient.

Question 8

A sequence begins with the terms 12,17,22,27,…12, 17, 22, 27, \ldots. If the nnth term of this sequence is 252252, what is the value of nn?

  1. 49 (correct answer)
  2. 48
  3. 50
  4. 51
Explanation: This is an arithmetic sequence with first term a1=12a_1 = 12 and common difference d=5d = 5. The general formula is an=12+(n−1)⋅5=5n+7a_n = 12 + (n-1) \cdot 5 = 5n + 7. Setting 5n+7=2525n + 7 = 252, we get 5n=2455n = 245, so n=49n = 49. Choice B results from using an=5n+2a_n = 5n + 2 (incorrect first term calculation). Choice C comes from solving 5n=2505n = 250 (forgetting the constant term). Choice D results from an off-by-one error in the formula.

Question 9

An arithmetic sequence has 1515 terms. The first term is aa and the last term is bb. If the middle term (88th term) is 2323, what is the sum of the first and last terms?

  1. 46 (correct answer)
  2. 23
  3. 38
  4. Cannot be determined
Explanation: In an arithmetic sequence with nn terms, the middle term equals the average of the first and last terms. For 15 terms, the 8th term is the middle term. So a8=a+b2=23a_8 = \frac{a + b}{2} = 23, which means a+b=46a + b = 46. Choice B incorrectly thinks the sum equals the middle term. Choice C assumes some specific values for aa and bb. Choice D incorrectly believes more information is needed, missing the key property of arithmetic sequences.