Arithmetic Sequences - Math 1

Consider the arithmetic sequence $2, 7, 12, 17, 22, ...$ 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?
Answer: $157 = 2 + 5(n-1)$ where $n \geq 1$ The arithmetic sequence has first term $a_1 = 2$ and common difference $d = 5$ . The general term is $a_n = 2 + (n-1) \cdot 5$ where $n \geq 1$ represents the position in the sequence. To check if 157 is a term, we need $157 = 2 + 5(n-1)$ for some positive integer $n$ . Solving: $155 = 5(n-1)$ , so $n-1 = 31$ and $n = 32$ . Since this gives an integer solution, 157 is indeed the 32nd term. Choice A uses $n \geq 0$ which would make the first term correspond to $n = 0$ , 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.
Consider the sequence $a_n = 5n - 2$ . Which statement about this sequence is NOT true?
Answer: The sequence contains both positive and negative terms For $a_n = 5n - 2$ : Choice A is true since $a_{n+1} - a_n = 5(n+1) - 2 - (5n - 2) = 5$ . Choice B is true since $a_1 = 5(1) - 2 = 3$ . Choice C is true since $a_{10} - a_7 = (48) - (33) = 15$ . Choice D is false because for $n \geq 1$ , we have $a_n = 5n - 2 \geq 5(1) - 2 = 3 > 0$ , so all terms are positive. Students might think negative terms exist by considering $n \leq 0$ , but sequence indices start at $n = 1$ .
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 $s$ seats and there are $n$ rows total, which expression represents the number of seats in the row that is exactly in the middle when $n$ is odd?
Answer: $s + 4\left(\frac{n-1}{2}\right)$ Since each row has 4 more seats than the previous row, this forms an arithmetic sequence with first term $s$ and common difference $d = 4$ . The general term is $a_k = s + (k-1) \cdot 4$ where $k$ is the row number. When $n$ is odd, the middle row is row number $\frac{n+1}{2}$ . So the number of seats in the middle row is $a_{\frac{n+1}{2}} = s + \left(\frac{n+1}{2} - 1\right) \cdot 4 = s + \left(\frac{n+1-2}{2}\right) \cdot 4 = s + 4\left(\frac{n-1}{2}\right)$ . Choice B incorrectly uses $\frac{n+1}{2}$ 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.
An arithmetic sequence has $15$ terms. The first term is $a$ and the last term is $b$ . If the middle term ( $8$ th term) is $23$ , what is the sum of the first and last terms?
Answer: 46 In an arithmetic sequence with $n$ terms, the middle term equals the average of the first and last terms. For 15 terms, the 8th term is the middle term. So $a_8 = \frac{a + b}{2} = 23$ , which means $a + b = 46$ . Choice B incorrectly thinks the sum equals the middle term. Choice C assumes some specific values for $a$ and $b$ . Choice D incorrectly believes more information is needed, missing the key property of arithmetic sequences.
An arithmetic sequence has first term $a$ and common difference $d > 0$ . If exactly three terms of this sequence are two-digit numbers (from 10 to 99), which of the following relationships must be true?
Answer: $a \leq 97$ and $d > 45$ 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 $a_n, a_{n+1}, a_{n+2}$ where $10 \leq a_n, a_{n+1}, a_{n+2} \leq 99$ but $a_{n-1} < 10$ (if it exists) and $a_{n+3} > 99$ . This means: $a_{n+2} \leq 99 < a_{n+3} = a_{n+2} + d$ . So $a_{n+2} \leq 99$ and $a_{n+2} + d > 99$ , which gives us $99 - d < a_{n+2} \leq 99$ . Since $a_{n+2} = a_n + 2d$ and $a_n \geq 10$ , we have $a_n + 2d \leq 99$ , so $a_n \leq 99 - 2d$ . Also, $a_n + 2d > 99 - d$ , so $a_n > 99 - 3d$ . For the sequence to have exactly three two-digit terms, we need $a_n \geq 10$ but $a_{n-1} = a_n - d < 10$ , giving us $10 - d < a_n < 10 + d$ (approximately). The key insight is that $d$ 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 $10 + 2d \leq 99$ and $10 + 3d > 99$ . This gives $d \leq 44.5$ and $d > 29.67$ . But this contradicts choice D. Let me reconsider: if $a$ itself could be two-digit and we want exactly three terms total to be two-digit, then $a \leq 99$ , $a + d \leq 99$ , $a + 2d \leq 99$ , but $a + 3d > 99$ . The last condition gives $d > \frac{99-a}{3}$ . For this to be satisfied with $d > 45$ , we need $45 > \frac{99-a}{3}$ , so $135 > 99 - a$ , giving $a > -36$ , which is always true for reasonable values. The condition $a \leq 97$ ensures that when $d > 45$ , we can have exactly three terms in the two-digit range.
A sequence begins with the terms $12, 17, 22, 27, \ldots$ . If the $n$ th term of this sequence is $252$ , what is the value of $n$ ?
Answer: 49 This is an arithmetic sequence with first term $a_1 = 12$ and common difference $d = 5$ . The general formula is $a_n = 12 + (n-1) \cdot 5 = 5n + 7$ . Setting $5n + 7 = 252$ , we get $5n = 245$ , so $n = 49$ . Choice B results from using $a_n = 5n + 2$ (incorrect first term calculation). Choice C comes from solving $5n = 250$ (forgetting the constant term). Choice D results from an off-by-one error in the formula.
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?
Answer: 27 Sequence A: $a_n^{(A)} = 8 + (n-1) \cdot 3 = 3n + 5$ . Sequence B: $a_n^{(B)} = 8 + (n-1) \cdot 5 = 5n + 3$ . We want the smallest $n$ such that $a_n^{(B)} - a_n^{(A)} > 50$ . This gives us $(5n + 3) - (3n + 5) > 50$ , which simplifies to $2n - 2 > 50$ , so $2n > 52$ , and $n > 26$ . Therefore, $n = 27$ is the first integer value where sequence B exceeds sequence A by more than 50. Let's verify: $a_{27}^{(B)} = 5(27) + 3 = 138$ and $a_{27}^{(A)} = 3(27) + 5 = 86$ . The difference is $138 - 86 = 52 > 50$ ✓. For $n = 26$ : $a_{26}^{(B)} = 133$ and $a_{26}^{(A)} = 83$ , difference = 50, which is not greater than 50.
An arithmetic sequence has the property that when each term is squared, the resulting sequence is $4, 25, 64, 121, 196, \ldots$ . What is the common difference of the original arithmetic sequence?
Answer: 3 or -3 The squared sequence is $4, 25, 64, 121, 196, \ldots = 2^2, 5^2, 8^2, 11^2, 14^2, \ldots$ . This means the original sequence could be $2, 5, 8, 11, 14, \ldots$ (with common difference $3$ ) or $-2, -5, -8, -11, -14, \ldots$ (with common difference $-3$ ). 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.
Two arithmetic sequences have the same first term. The first sequence has a common difference of $3$ and its $8$ th term is $29$ . The second sequence has a common difference of $-2$ and its $k$ th term equals the $12$ th term of the first sequence. What is the value of $k$ ?
Answer: 19 For the first sequence: $a_8 = a_1 + 7(3) = 29$ , so $a_1 = 8$ . The 12th term is $a_{12} = 8 + 11(3) = 41$ . For the second sequence with $a_1 = 8$ and $d = -2$ : $a_k = 8 + (k-1)(-2) = 41$ . Solving: $8 - 2k + 2 = 41$ , so $10 - 2k = 41$ , giving $k = 19$ . Choice A uses $k-1 = 17$ 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.

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