Arithmetic Sequences & Linear Functions - Math 1

An arithmetic sequence has the property that when its terms are plotted as points $(n, a_n)$ on a coordinate plane, they lie on the line $y = -2x + 16$ . What is the sum of the first 6 terms of this arithmetic sequence?
Answer: $54$ Since the terms lie on $y = -2x + 16$ , we have $a_n = -2n + 16$ . The first 6 terms are: $a_1 = 14$ , $a_2 = 12$ , $a_3 = 10$ , $a_4 = 8$ , $a_5 = 6$ , $a_6 = 4$ . The sum is $14 + 12 + 10 + 8 + 6 + 4 = 54$ . 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.
Two linear functions $p(x) = 2x + 5$ and $q(x) = -x + 14$ generate arithmetic sequences when evaluated at consecutive positive integers. At what positive integer value of $x$ do these functions produce the same output, and what is that output?
Answer: $x = 3$ , output = $11$ Setting $p(x) = q(x)$ : $2x + 5 = -x + 14$ , so $3x = 9$ , giving $x = 3$ . At $x = 3$ : $p(3) = 2(3) + 5 = 11$ and $q(3) = -3 + 14 = 11$ . Choice B uses $x = 4$ incorrectly. Choice C uses $x = 2$ . Choice D uses $x = 5$ , all giving wrong intersection points.
Two linear functions $g(x) = 3x + 7$ and $h(x) = 5x - 1$ each generate arithmetic sequences when evaluated at positive integers. For what positive integer value of $x$ is the difference $h(x) - g(x)$ equal to 18?
Answer: $x = 13$ We need $h(x) - g(x) = 18$ , so $(5x - 1) - (3x + 7) = 18$ . Simplifying: $5x - 1 - 3x - 7 = 18$ , which gives $2x - 8 = 18$ , so $2x = 26$ and $x = 13$ . Choice A results from solving $2x - 8 = 14$ . Choice B comes from calculation errors in the algebraic manipulation. Choice D results from solving $2x - 8 = 20$ .
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 $f(n)$ , what is the value of $f(20)$ ?
Answer: $68$ Given $a_5 = 23$ and $a_{12} = 44$ . The common difference is $d = \frac{44 - 23}{12 - 5} = \frac{21}{7} = 3$ . Using $a_5 = 23$ : $a_1 + 4d = 23$ , so $a_1 + 12 = 23$ , giving $a_1 = 11$ . The general term is $a_n = 11 + (n-1)(3) = 3n + 8$ . Therefore $f(20) = 3(20) + 8 = 68$ . 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.
An arithmetic sequence can be represented by the linear function $f(n) = 7n - 3$ . If we consider only the terms where $n$ is a multiple of 3 (i.e., $n = 3, 6, 9, 12, ...$ ), what is the common difference of this new sequence?
Answer: $21$ When $n$ is a multiple of 3, we evaluate $f(3) = 18$ , $f(6) = 39$ , $f(9) = 60$ , $f(12) = 81$ , etc. The differences are $39 - 18 = 21$ , $60 - 39 = 21$ , $81 - 60 = 21$ . 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 $3 \times 7 = 21$ . Choice A is the original common difference. Choice B results from taking every 2nd term. Choice D results from calculation errors.
An arithmetic sequence can be written as $a_n = 15 - 4n$ . 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?
Answer: $-4$ , representing the common difference When graphing an arithmetic sequence as a linear function, the slope equals the common difference. The formula $a_n = 15 - 4n$ shows the slope is $-4$ , 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 $a_1 = 15 - 4(1) = 11$ but this is the first term, not the slope.
Consider the arithmetic sequence $5, 11, 17, 23, ...$ 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?
Answer: $-1$ The arithmetic sequence $5, 11, 17, 23, ...$ has first term $a_1 = 5$ and common difference $d = 6$ . The general term is $a_n = 5 + (n-1)(6) = 6n - 1$ . The 0th term would be $a_0 = 6(0) - 1 = -1$ . 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.
An arithmetic sequence has first term $a_1 = 7$ and common difference $d = -3$ . If this sequence is represented as a linear function $f(n) = an + b$ where $n$ represents the term position, what is the value of $a + b$ ?
Answer: $7$ For an arithmetic sequence with first term $a_1 = 7$ and common difference $d = -3$ , the general term is $a_n = 7 + (n-1)(-3) = 7 - 3n + 3 = 10 - 3n$ . So $f(n) = -3n + 10$ , meaning $a = -3$ and $b = 10$ . Therefore $a + b = -3 + 10 = 7$ . Choice A results from confusing the slope with the y-intercept. Choice B comes from using $d$ as the coefficient directly without the negative sign. Choice D results from just taking the y-intercept value.
The linear function $g(x) = 4x - 5$ generates an arithmetic sequence when evaluated at consecutive positive integers. What is the sum of the 8th and 12th terms of this arithmetic sequence?
Answer: $66$ When $g(x) = 4x - 5$ is evaluated at consecutive positive integers, we get an arithmetic sequence. The 8th term is $g(8) = 4(8) - 5 = 27$ and the 12th term is $g(12) = 4(12) - 5 = 39$ . The sum is $27 + 39 = 66$ . Choice A results from calculation errors in function evaluation. Choice B comes from finding $g(7) + g(12)$ instead. Choice C results from finding $g(8) + g(11)$ instead.
A linear function $r(t) = mt + b$ models an arithmetic sequence where consecutive integer inputs produce consecutive terms. If $r(3) = 17$ and $r(7) = 29$ , what is the value of $r(0)$ and what does it represent in the context of the arithmetic sequence?
Answer: $r(0) = 8$ ; this is the theoretical zeroth term 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: $m = \frac{29 - 17}{7 - 3} = \frac{12}{4} = 3$ . This means each term increases by 3. Now use point-slope form with either given point. Using $(3, 17)$ : $r(t) - 17 = 3(t - 3)$ , which simplifies to $r(t) = 3t + 8$ . Therefore, $r(0) = 3(0) + 8 = 8$ . In the context of an arithmetic sequence, $r(0)$ 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 $r(0)$ 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 $r(0)$ gives you the y-intercept, which represents the starting point when the input is zero.
The linear function $k(t) = -2t + 64$ generates an arithmetic sequence. If the sum of three consecutive terms of this sequence is 120, and these terms correspond to $k(n)$ , $k(n+1)$ , and $k(n+2)$ for some positive integer $n$ , what is the value of $n$ ?
Answer: $n = 11$ When you see a linear function generating an arithmetic sequence, remember that consecutive terms will have a constant difference. Here, $k(t) = -2t + 64$ has a common difference of $-2$ . The three consecutive terms are $k(n)$ , $k(n+1)$ , and $k(n+2)$ . Let's express each: - $k(n) = -2n + 64$ - $k(n+1) = -2(n+1) + 64 = -2n + 62$ - $k(n+2) = -2(n+2) + 64 = -2n + 60$ Since their sum equals 120: $(-2n + 64) + (-2n + 62) + (-2n + 60) = 120$ $-6n + 186 = 120$ $-6n = -66$ $n = 11$ Let's verify: $k(11) = 42$ , $k(12) = 40$ , $k(13) = 38$ , and $42 + 40 + 38 = 120$ ✓ Now examining the wrong answers: Choice B ( $n = 10$ ) gives us terms 44, 42, 40 with sum 126 – too large. Choice C ( $n = 12$ ) yields terms 40, 38, 36 with sum 114 – too small. Choice D ( $n = 13$ ) produces terms 38, 36, 34 with sum 108 – also too small. Notice that as $n$ 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.
An arithmetic sequence has its first term $a_1 = 7$ and common difference $d = -3$ . If this sequence is represented as a linear function $f(n) = mn + b$ where $n$ represents the position number, what is the value of $f(0)$ ?
Answer: $10$ For an arithmetic sequence $a_n = a_1 + (n-1)d = 7 + (n-1)(-3) = 7 - 3n + 3 = 10 - 3n$ . To express this as $f(n) = mn + b$ , we have $f(n) = -3n + 10$ , so $f(0) = -3(0) + 10 = 10$ . Choice A results from using $a_1$ incorrectly. Choice B assumes $f(0) = a_1$ . Choice D uses the common difference instead of the y-intercept.
A linear function $p(x) = mx + c$ passes through the points $(1, 12)$ and $(4, 27)$ . When this function generates an arithmetic sequence by evaluating $p(1), p(2), p(3), ...$ , what is the 15th term of this sequence?
Answer: $82$ Using the two points $(1, 12)$ and $(4, 27)$ , the slope is $m = \frac{27-12}{4-1} = \frac{15}{3} = 5$ . Using point-slope form with $(1, 12)$ : $12 = 5(1) + c$ , so $c = 7$ . Therefore $p(x) = 5x + 7$ . The 15th term is $p(15) = 5(15) + 7 = 75 + 7 = 82$ . 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.
A linear function $g(x) = 4x - 5$ generates values that form an arithmetic sequence when $x$ takes consecutive integer values starting from $x = 1$ . What is the 8th term of this arithmetic sequence?
Answer: $23$ When $x$ takes consecutive integers starting from 1, we get: $g(1) = 4(1) - 5 = -1$ , $g(2) = 4(2) - 5 = 3$ , $g(3) = 4(3) - 5 = 7$ , etc. This forms the arithmetic sequence $-1, 3, 7, 11, ...$ with first term $a_1 = -1$ and common difference $d = 4$ . The 8th term is $a_8 = -1 + (8-1)(4) = -1 + 28 = 23$ . Choice A uses $x = 8$ directly in $g(x)$ . Choice C miscalculates the arithmetic. Choice D uses the wrong starting position.
An arithmetic sequence begins $2, 8, 14, 20, ...$ If this sequence corresponds to a linear function $h(t)$ where the domain represents term positions starting at $t = 1$ , what is the value of $h(0)$ ?
Answer: $-4$ The arithmetic sequence $2, 8, 14, 20, ...$ has first term $a_1 = 2$ and common difference $d = 6$ . The general term is $a_n = 2 + (n-1)(6) = 6n - 4$ . So $h(t) = 6t - 4$ . Therefore $h(0) = 6(0) - 4 = -4$ . Choice B results from using the wrong linear function form. Choice C comes from assuming $h(0) = 0$ . Choice D results from confusing $h(0)$ with the first term of the sequence.

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