Math 1 Quiz: Arithmetic Sequences And Linear Functions
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Arithmetic Sequences And Linear FunctionsQuestion 1 of 15

An arithmetic sequence has first term a1=7a_1 = 7 and common difference d=−3d = -3. If this sequence is represented as a linear function f(n)=an+bf(n) = an + b where nn represents the term position, what is the value of a+ba + b?

11
44
77
1010
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Math 1 Quiz

Math 1 Quiz: Arithmetic Sequences And Linear Functions

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

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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.

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Question 1

An arithmetic sequence has first term a1=7a_1 = 7 and common difference d=−3d = -3. If this sequence is represented as a linear function f(n)=an+bf(n) = an + b where nn represents the term position, what is the value of a+ba + b?

  1. 11
  2. 44
  3. 77 (correct answer)
  4. 1010
Explanation: For an arithmetic sequence with first term a1=7a_1 = 7 and common difference d=−3d = -3, the general term is an=7+(n−1)(−3)=7−3n+3=10−3na_n = 7 + (n-1)(-3) = 7 - 3n + 3 = 10 - 3n. So f(n)=−3n+10f(n) = -3n + 10, meaning a=−3a = -3 and b=10b = 10. Therefore a+b=−3+10=7a + b = -3 + 10 = 7. Choice A results from confusing the slope with the y-intercept. Choice B comes from using dd as the coefficient directly without the negative sign. Choice D results from just taking the y-intercept value.

Question 2

A linear function p(x)=mx+cp(x) = mx + c passes through the points (1,12)(1, 12) and (4,27)(4, 27). When this function generates an arithmetic sequence by evaluating p(1),p(2),p(3),...p(1), p(2), p(3), ..., what is the 15th term of this sequence?

  1. 7272
  2. 7575
  3. 7777
  4. 8282 (correct answer)
Explanation: Using the two points (1,12)(1, 12) and (4,27)(4, 27), the slope is m=27−124−1=153=5m = \frac{27-12}{4-1} = \frac{15}{3} = 5. Using point-slope form with (1,12)(1, 12): 12=5(1)+c12 = 5(1) + c, so c=7c = 7. Therefore p(x)=5x+7p(x) = 5x + 7. The 15th term is p(15)=5(15)+7=75+7=82p(15) = 5(15) + 7 = 75 + 7 = 82. Choice A results from calculation errors. Choice B comes from forgetting to add the constant term. Choice C results from using an incorrect y-intercept value.

Question 3

An arithmetic sequence can be represented by the linear function f(n)=7n−3f(n) = 7n - 3. If we consider only the terms where nn is a multiple of 3 (i.e., n=3,6,9,12,...n = 3, 6, 9, 12, ...), what is the common difference of this new sequence?

  1. 77
  2. 1414
  3. 2121 (correct answer)
  4. 2828
Explanation: When nn is a multiple of 3, we evaluate f(3)=18f(3) = 18, f(6)=39f(6) = 39, f(9)=60f(9) = 60, f(12)=81f(12) = 81, etc. The differences are 39−18=2139 - 18 = 21, 60−39=2160 - 39 = 21, 81−60=2181 - 60 = 21. So the common difference is 21. This makes sense because we're taking every 3rd term of the original sequence, so the new common difference is 3×7=213 \times 7 = 21. Choice A is the original common difference. Choice B results from taking every 2nd term. Choice D results from calculation errors.

Question 4

An arithmetic sequence begins 2,8,14,20,...2, 8, 14, 20, ... If this sequence corresponds to a linear function h(t)h(t) where the domain represents term positions starting at t=1t = 1, what is the value of h(0)h(0)?

  1. −4-4 (correct answer)
  2. −2-2
  3. 00
  4. 22
Explanation: The arithmetic sequence 2,8,14,20,...2, 8, 14, 20, ... has first term a1=2a_1 = 2 and common difference d=6d = 6. The general term is an=2+(n−1)(6)=6n−4a_n = 2 + (n-1)(6) = 6n - 4. So h(t)=6t−4h(t) = 6t - 4. Therefore h(0)=6(0)−4=−4h(0) = 6(0) - 4 = -4. Choice B results from using the wrong linear function form. Choice C comes from assuming h(0)=0h(0) = 0. Choice D results from confusing h(0)h(0) with the first term of the sequence.

Question 5

The linear function k(t)=−2t+64k(t) = -2t + 64 generates an arithmetic sequence. If the sum of three consecutive terms of this sequence is 120, and these terms correspond to k(n)k(n), k(n+1)k(n+1), and k(n+2)k(n+2) for some positive integer nn, what is the value of nn?

  1. n=11n = 11 (correct answer)
  2. n=10n = 10
  3. n=12n = 12
  4. n=13n = 13
Explanation: When you see a linear function generating an arithmetic sequence, remember that consecutive terms will have a constant difference. Here, k(t)=−2t+64k(t) = -2t + 64 has a common difference of −2-2. The three consecutive terms are k(n)k(n), k(n+1)k(n+1), and k(n+2)k(n+2). Let's express each:
  • k(n)=−2n+64k(n) = -2n + 64
  • k(n+1)=−2(n+1)+64=−2n+62k(n+1) = -2(n+1) + 64 = -2n + 62
  • k(n+2)=−2(n+2)+64=−2n+60k(n+2) = -2(n+2) + 64 = -2n + 60
Since their sum equals 120: (−2n+64)+(−2n+62)+(−2n+60)=120(-2n + 64) + (-2n + 62) + (-2n + 60) = 120 −6n+186=120-6n + 186 = 120 −6n=−66-6n = -66 n=11n = 11 Let's verify: k(11)=42k(11) = 42, k(12)=40k(12) = 40, k(13)=38k(13) = 38, and 42+40+38=12042 + 40 + 38 = 120 ✓ Now examining the wrong answers: Choice B (n=10n = 10) gives us terms 44, 42, 40 with sum 126 – too large. Choice C (n=12n = 12) yields terms 40, 38, 36 with sum 114 – too small. Choice D (n=13n = 13) produces terms 38, 36, 34 with sum 108 – also too small. Notice that as nn increases, the sum decreases because this is a decreasing linear function (negative slope). Each wrong answer represents a systematic shift from the correct starting position. Study tip: When working with arithmetic sequences from linear functions, set up your equation by adding the explicit terms rather than using general formulas. This reduces algebraic errors and makes verification straightforward.

Question 6

Two linear functions g(x)=3x+7g(x) = 3x + 7 and h(x)=5x−1h(x) = 5x - 1 each generate arithmetic sequences when evaluated at positive integers. For what positive integer value of xx is the difference h(x)−g(x)h(x) - g(x) equal to 18?

  1. x=11x = 11
  2. x=12x = 12
  3. x=13x = 13 (correct answer)
  4. x=14x = 14
Explanation: We need h(x)−g(x)=18h(x) - g(x) = 18, so (5x−1)−(3x+7)=18(5x - 1) - (3x + 7) = 18. Simplifying: 5x−1−3x−7=185x - 1 - 3x - 7 = 18, which gives 2x−8=182x - 8 = 18, so 2x=262x = 26 and x=13x = 13. Choice A results from solving 2x−8=142x - 8 = 14. Choice B comes from calculation errors in the algebraic manipulation. Choice D results from solving 2x−8=202x - 8 = 20.

Question 7

An arithmetic sequence has the property that when its terms are plotted as points (n,an)(n, a_n) on a coordinate plane, they lie on the line y=−2x+16y = -2x + 16. What is the sum of the first 6 terms of this arithmetic sequence?

  1. 4848
  2. 5454 (correct answer)
  3. 6060
  4. 6666
Explanation: Since the terms lie on y=−2x+16y = -2x + 16, we have an=−2n+16a_n = -2n + 16. The first 6 terms are: a1=14a_1 = 14, a2=12a_2 = 12, a3=10a_3 = 10, a4=8a_4 = 8, a5=6a_5 = 6, a6=4a_6 = 4. The sum is 14+12+10+8+6+4=5414 + 12 + 10 + 8 + 6 + 4 = 54. Choice A results from using the wrong y-intercept. Choice C comes from calculation errors in the sum. Choice D results from using an incorrect common difference.

Question 8

An arithmetic sequence has its first term a1=7a_1 = 7 and common difference d=−3d = -3. If this sequence is represented as a linear function f(n)=mn+bf(n) = mn + b where nn represents the position number, what is the value of f(0)f(0)?

  1. 44
  2. 77
  3. 1010 (correct answer)
  4. −3-3
Explanation: For an arithmetic sequence an=a1+(n−1)d=7+(n−1)(−3)=7−3n+3=10−3na_n = a_1 + (n-1)d = 7 + (n-1)(-3) = 7 - 3n + 3 = 10 - 3n. To express this as f(n)=mn+bf(n) = mn + b, we have f(n)=−3n+10f(n) = -3n + 10, so f(0)=−3(0)+10=10f(0) = -3(0) + 10 = 10. Choice A results from using a1a_1 incorrectly. Choice B assumes f(0)=a1f(0) = a_1. Choice D uses the common difference instead of the y-intercept.

Question 9

A linear function g(x)=4x−5g(x) = 4x - 5 generates values that form an arithmetic sequence when xx takes consecutive integer values starting from x=1x = 1. What is the 8th term of this arithmetic sequence?

  1. 2727
  2. 2323 (correct answer)
  3. 3131
  4. 1919
Explanation: When xx takes consecutive integers starting from 1, we get: g(1)=4(1)−5=−1g(1) = 4(1) - 5 = -1, g(2)=4(2)−5=3g(2) = 4(2) - 5 = 3, g(3)=4(3)−5=7g(3) = 4(3) - 5 = 7, etc. This forms the arithmetic sequence −1,3,7,11,...-1, 3, 7, 11, ... with first term a1=−1a_1 = -1 and common difference d=4d = 4. The 8th term is a8=−1+(8−1)(4)=−1+28=23a_8 = -1 + (8-1)(4) = -1 + 28 = 23. Choice A uses x=8x = 8 directly in g(x)g(x). Choice C miscalculates the arithmetic. Choice D uses the wrong starting position.

Question 10

Two linear functions p(x)=2x+5p(x) = 2x + 5 and q(x)=−x+14q(x) = -x + 14 generate arithmetic sequences when evaluated at consecutive positive integers. At what positive integer value of xx do these functions produce the same output, and what is that output?

  1. x=3x = 3, output = 1111 (correct answer)
  2. x=4x = 4, output = 1010
  3. x=2x = 2, output = 1212
  4. x=5x = 5, output = 99
Explanation: Setting p(x)=q(x)p(x) = q(x): 2x+5=−x+142x + 5 = -x + 14, so 3x=93x = 9, giving x=3x = 3. At x=3x = 3: p(3)=2(3)+5=11p(3) = 2(3) + 5 = 11 and q(3)=−3+14=11q(3) = -3 + 14 = 11. Choice B uses x=4x = 4 incorrectly. Choice C uses x=2x = 2. Choice D uses x=5x = 5, all giving wrong intersection points.

Question 11

A linear function r(t)=mt+br(t) = mt + b models an arithmetic sequence where consecutive integer inputs produce consecutive terms. If r(3)=17r(3) = 17 and r(7)=29r(7) = 29, what is the value of r(0)r(0) and what does it represent in the context of the arithmetic sequence?

  1. r(0)=2r(0) = 2; this is 3 terms before the first term
  2. r(0)=5r(0) = 5; this is 3 terms before the first term
  3. r(0)=11r(0) = 11; this is the theoretical zeroth term
  4. r(0)=8r(0) = 8; this is the theoretical zeroth term (correct answer)
Explanation: When you see a linear function modeling an arithmetic sequence, remember that the slope represents the common difference between consecutive terms, and you can find any term using the two-point form of a line. First, find the slope (common difference) using the given points: m=29−177−3=124=3m = \frac{29 - 17}{7 - 3} = \frac{12}{4} = 3. This means each term increases by 3. Now use point-slope form with either given point. Using (3,17)(3, 17): r(t)−17=3(t−3)r(t) - 17 = 3(t - 3), which simplifies to r(t)=3t+8r(t) = 3t + 8. Therefore, r(0)=3(0)+8=8r(0) = 3(0) + 8 = 8. In the context of an arithmetic sequence, r(0)r(0) represents the theoretical "zeroth term" - the value that would come immediately before the first term if we extended the pattern backward. Choice A gives the wrong value (2 instead of 8) and incorrectly describes r(0)r(0) as "3 terms before the first term." Choice B also has the wrong value (5) with the same incorrect description. Choice C gets closer with the right interpretation ("theoretical zeroth term") but calculates the wrong value (11 instead of 8). This likely comes from arithmetic errors in the slope calculation or point-slope substitution. Choice D correctly identifies both the value (8) and the meaning (theoretical zeroth term). Study tip: When working with linear functions and arithmetic sequences, always double-check your slope calculation first - it's the foundation for everything else. Remember that r(0)r(0) gives you the y-intercept, which represents the starting point when the input is zero.

Question 12

An arithmetic sequence can be written as an=15−4na_n = 15 - 4n. When this sequence is graphed as a linear function with the term position on the x-axis and the term value on the y-axis, what is the slope of the resulting line?

  1. −4-4, representing the common difference (correct answer)
  2. 44, representing the absolute common difference
  3. 1515, representing the first term value
  4. 1111, representing the first term a1a_1
Explanation: When graphing an arithmetic sequence as a linear function, the slope equals the common difference. The formula an=15−4na_n = 15 - 4n shows the slope is −4-4, which is also the common difference of the sequence. Choice B incorrectly takes the absolute value. Choice C confuses the y-intercept with slope. Choice D calculates a1=15−4(1)=11a_1 = 15 - 4(1) = 11 but this is the first term, not the slope.

Question 13

Consider the arithmetic sequence 5,11,17,23,...5, 11, 17, 23, ... If this sequence is extended to include a "0th term" by following the same linear pattern, what would be the value of this 0th term?

  1. −7-7
  2. −1-1 (correct answer)
  3. 22
  4. 55
Explanation: The arithmetic sequence 5,11,17,23,...5, 11, 17, 23, ... has first term a1=5a_1 = 5 and common difference d=6d = 6. The general term is an=5+(n−1)(6)=6n−1a_n = 5 + (n-1)(6) = 6n - 1. The 0th term would be a0=6(0)−1=−1a_0 = 6(0) - 1 = -1. Choice A results from miscalculating the linear function. Choice C comes from arithmetic errors. Choice D results from confusing the 0th term with the 1st term.

Question 14

The linear function g(x)=4x−5g(x) = 4x - 5 generates an arithmetic sequence when evaluated at consecutive positive integers. What is the sum of the 8th and 12th terms of this arithmetic sequence?

  1. 5454
  2. 5858
  3. 6262
  4. 6666 (correct answer)
Explanation: When g(x)=4x−5g(x) = 4x - 5 is evaluated at consecutive positive integers, we get an arithmetic sequence. The 8th term is g(8)=4(8)−5=27g(8) = 4(8) - 5 = 27 and the 12th term is g(12)=4(12)−5=39g(12) = 4(12) - 5 = 39. The sum is 27+39=6627 + 39 = 66. Choice A results from calculation errors in function evaluation. Choice B comes from finding g(7)+g(12)g(7) + g(12) instead. Choice C results from finding g(8)+g(11)g(8) + g(11) instead.

Question 15

An arithmetic sequence has its 5th term equal to 23 and its 12th term equal to 44. If this sequence is represented as a linear function f(n)f(n), what is the value of f(20)f(20)?

  1. 6565
  2. 6868 (correct answer)
  3. 7171
  4. 7474
Explanation: Given a5=23a_5 = 23 and a12=44a_{12} = 44. The common difference is d=44−2312−5=217=3d = \frac{44 - 23}{12 - 5} = \frac{21}{7} = 3. Using a5=23a_5 = 23: a1+4d=23a_1 + 4d = 23, so a1+12=23a_1 + 12 = 23, giving a1=11a_1 = 11. The general term is an=11+(n−1)(3)=3n+8a_n = 11 + (n-1)(3) = 3n + 8. Therefore f(20)=3(20)+8=68f(20) = 3(20) + 8 = 68. Choice A results from using the wrong first term. Choice C comes from calculation errors in finding the common difference. Choice D results from arithmetic errors in the final computation.