Recursive Rules for Sequences - Math 1

A student writes two different recursive rules for arithmetic sequences: Rule 1: $f_1 = 5, f_n = f_{n-1} + 3$ and Rule 2: $g_1 = 5, g_n = g_{n-1} + 3$ . The student claims these generate different sequences because they use different variable names. How should you respond to this claim?
Answer: The claim is incorrect because variable names are arbitrary labels and both rules generate identical sequences When you encounter questions about mathematical notation and variable names, focus on the fundamental principle that mathematics cares about relationships and patterns, not the specific symbols we use to represent them. Both recursive rules describe exactly the same arithmetic sequence. Rule 1 generates: $f_1 = 5, f_2 = 8, f_3 = 11, f_4 = 14, ...$ Rule 2 generates: $g_1 = 5, g_2 = 8, g_3 = 11, g_4 = 14, ...$ The sequences are identical because both start with 5 and add 3 at each step. Variable names like $f_n$ and $g_n$ are simply labels—they're like calling the same person "John" versus "Jonathan." The mathematical relationship remains unchanged. Choice A is incorrect because $f_n$ and $g_n$ represent the same mathematical object despite different notation. Choice B misses the point entirely—both sequences have the same domain (positive integers) and the same range values. Choice D incorrectly suggests the initial conditions differ, but both rules explicitly state the first term equals 5. Choice C correctly identifies that variable names are arbitrary labels that don't affect the underlying mathematical relationship. When the initial value and recursive rule are identical, the sequences must be identical regardless of notation. Remember this key principle: In mathematics, substance matters more than symbols. Two expressions or sequences are equivalent if they produce the same results, even when written with different variable names. Always look past the notation to the underlying mathematical relationships.
An arithmetic sequence has the recursive rule $c_1 = -6$ and $c_n = c_{n-1} + d$ where $d$ is the common difference. If $c_7 = 18$ , what recursive rule completely describes this sequence?
Answer: $c_1 = -6$ , $c_n = c_{n-1} + 4$ for $n \geq 2$ From $c_1 = -6$ to $c_7 = 18$ requires 6 steps of adding the common difference $d$ . So $-6 + 6d = 18$ , which gives $6d = 24$ , thus $d = 4$ . The complete recursive rule is $c_1 = -6$ , $c_n = c_{n-1} + 4$ for $n \geq 2$ . Choice B uses the total change (24) instead of the common difference. Choice C reverses the first and seventh terms. Choice D incorrectly starts the recursion at $n = 8$ .
A student claims that $w_1 = 8$ , $w_n = 2w_{n-1} - 8$ is a valid recursive rule for an arithmetic sequence. Which analysis of this claim is correct?
Answer: The claim is correct because the rule produces the constant sequence 8, 8, 8, 8, ... with common difference 0 When evaluating whether a recursive formula generates an arithmetic sequence, you need to check if it produces terms with a constant common difference. The key insight is to actually compute the terms and examine the pattern. Let's work through the given recursion $w_1 = 8$ and $w_n = 2w_{n-1} - 8$ : - $w_1 = 8$ - $w_2 = 2(8) - 8 = 16 - 8 = 8$ - $w_3 = 2(8) - 8 = 16 - 8 = 8$ - $w_4 = 2(8) - 8 = 16 - 8 = 8$ The sequence is 8, 8, 8, 8, ... with a common difference of 0. Since the difference between consecutive terms is constant (zero), this is indeed an arithmetic sequence. Choice A incorrectly assumes the pattern will break down, but our calculation shows it continues indefinitely. Choice B makes the false claim that multiplication cannot generate arithmetic sequences—this is wrong because the specific combination of operations matters, not just the presence of multiplication. Choice C attempts a clever algebraic manipulation, but $w_n = w_{n-1} + (w_{n-1} - 8)$ doesn't show constant differences since $(w_{n-1} - 8)$ varies with the term value. Choice D correctly identifies that this recursion produces a constant sequence with common difference 0. Remember that arithmetic sequences are defined by having a constant common difference—this difference can be positive, negative, or zero. Don't let the presence of multiplication in a recursive formula fool you; always compute the actual terms to verify the pattern.
An arithmetic sequence has recursive rule $t_1 = a$ and $t_n = t_{n-1} + 7$ for $n \geq 2$ . If $t_4 - t_2 = 14$ , which statement about the value of $a$ is correct?
Answer: The value of $a$ can be any real number and the condition will still hold true In any arithmetic sequence with common difference 7, we have $t_4 = t_2 + 2(7) = t_2 + 14$ , so $t_4 - t_2 = 14$ regardless of the first term $a$ . Specifically: $t_2 = a + 7$ , $t_4 = a + 21$ , so $t_4 - t_2 = (a + 21) - (a + 7) = 14$ . The value of $a$ cancels out. Choices A and B incorrectly specify particular values. Choice D incorrectly restricts $a$ to positive values.
A sequence is defined by $a_1 = -3$ and $a_n = a_{n-1} + 5$ for $n \geq 2$ . If this pattern continues, what is the value of $a_{10} - a_7$ ?
Answer: 15 In an arithmetic sequence with common difference d = 5, the difference between any two terms separated by k positions is k × d. Since a₁₀ is 3 positions after a₇, we have a₁₀ - a₇ = 3 × 5 = 15. This can be verified by noting that a₇ = -3 + 6(5) = 27 and a₁₀ = -3 + 9(5) = 42, so a₁₀ - a₇ = 42 - 27 = 15. Choice B uses k = 2 instead of 3. Choice C uses the wrong common difference. Choice D miscalculates the position difference.
An arithmetic sequence has $a_1 = 7$ and $a_5 = 23$ . Maria writes the recursive rule as $a_n = a_{n-1} + 4$ for $n \geq 2$ . Which statement best explains why Maria's rule is incomplete?
Answer: The rule is missing the initial term $a_1 = 7$ which is required to generate the sequence A recursive rule for a sequence requires both the recursive relationship AND the initial condition. Maria correctly found the common difference (d = 4), but her rule is incomplete because it doesn't specify the starting value a₁ = 7. Choice B is wrong because d = (23-7)/(5-1) = 4. Choice C is wrong because n ≥ 2 is correct for the recursive part. Choice D is wrong because both forms of the recursive relationship are mathematically equivalent.
An arithmetic sequence has first term $a_1 = 7$ and common difference $d = -3$ . Which recursive rule correctly describes this sequence and produces $a_4 = -2$ ?
Answer: $a_1 = 7$ , $a_n = a_{n-1} - 3$ for $n \geq 2$ For an arithmetic sequence with first term 7 and common difference -3, the recursive rule is $a_1 = 7$ and $a_n = a_{n-1} + (-3) = a_{n-1} - 3$ for $n \geq 2$ . This gives: $a_1 = 7$ , $a_2 = 4$ , $a_3 = 1$ , $a_4 = -2$ . Choice B incorrectly references $a_{n+1}$ instead of $a_{n-1}$ . Choice C confuses the first term with the common difference. Choice D uses the wrong sign for the common difference.
A recursive rule $m_1 = 15$ and $m_n = m_{n-1} - d$ where $d > 0$ generates an arithmetic sequence. If $m_5 = 3$ , what is the recursive rule and what does it tell us about the sequence behavior?
Answer: $m_1 = 15$ , $m_n = m_{n-1} - 3$ for $n \geq 2$ ; the sequence decreases and will eventually become negative From $m_1 = 15$ to $m_5 = 3$ requires 4 steps: $15 - 4d = 3$ , so $4d = 12$ and $d = 3$ . The rule is $m_1 = 15$ , $m_n = m_{n-1} - 3$ . Since we subtract 3 each step, the sequence decreases and will eventually become negative ( $m_6 = 0$ , $m_7 = -3$ , etc.). Choice A has wrong behavior analysis. Choice C uses total change instead of common difference. Choice D has wrong sign and behavior.
Consider the recursive rule $b_1 = 8$ and $b_n = b_{n-1} + 6$ for $n \geq 2$ . Which statement about this sequence is correct?
Answer: The sequence has a constant rate of change of 6 units per term position This is an arithmetic sequence with common difference 6, meaning it has a constant rate of change of 6 units per term position. The sequence is: 8, 14, 20, 26, ... Choice A describes geometric sequences, not arithmetic. Choice C is wrong because the sequence increases indefinitely (8, 14, 20, ...). Choice D is wrong because arithmetic sequences represent linear relationships, not quadratic.
Two students write recursive rules for the same arithmetic sequence. Student A writes: $a_1 = 12, a_n = a_{n-1} - 3$ for $n \geq 2$ . Student B writes: $a_1 = 9, a_n = a_{n-1} - 3$ for $n \geq 2$ . If both rules generate valid arithmetic sequences with the same common difference, what can be concluded?
Answer: The sequences represent parallel patterns but Student A starts at a different position Both sequences have the same common difference (-3) but different starting values, creating parallel arithmetic sequences. Student A: 12, 9, 6, 3, ... and Student B: 9, 6, 3, 0, ... These are distinct but parallel patterns. Choice A is wrong because identical sequences must have the same terms. Choice B is wrong because the difference is 3 only for corresponding positions, not always. Choice D is wrong because multiple arithmetic sequences can have the same common difference.
Which recursive rule correctly models the number of seats in row $n$ ?
Answer: $s_1 = 18$ and $s_n = s_{n-1} + 4$ for $n \geq 2$ When you encounter a recursive sequence problem, you need to identify two key components: the starting value and the pattern that connects each term to the previous one. Let's work through this step-by-step. The theater's first row has 18 seats, and each subsequent row has 4 more seats than the previous row. This means: - Row 1: 18 seats - Row 2: 18 + 4 = 22 seats - Row 3: 22 + 4 = 26 seats - Row 4: 26 + 4 = 30 seats The recursive rule should capture this pattern: start with 18, then add 4 to get each next term. Choice D correctly models this: $s_1 = 18$ and $s_n = s_{n-1} + 4$ for $n \geq 2$ . This says the first row has 18 seats, and every subsequent row has 4 more seats than the row before it. Choice A is wrong because $s_n = s_{n-1} + 4n$ would add $4n$ (not just 4) each time, creating much larger jumps between rows. Choice B is incorrect because $s_n = 4s_{n-1}$ multiplies the previous term by 4 rather than adding 4, which would create an exponential growth pattern. Choice C fails because it starts with $s_1 = 22$ , but the problem clearly states the first row has 18 seats, not 22. Study tip: In recursive sequence problems, always check that your initial value matches the given information, and verify your recursive formula by calculating the first few terms to see if they match the described pattern.
A teacher asks students to write a recursive rule for an arithmetic sequence where the 10th term exceeds the 3rd term by 21. Which student response demonstrates the best understanding of the relationship between recursive rules and arithmetic sequences?
Answer: Student writes: The common difference must be 3, but the first term can be any value since it doesn't affect the given constraint The constraint $a_{10} - a_3 = 21$ means $7d = 21$ , so $d = 3$ . However, this constraint doesn't determine the first term—any value works as long as the common difference is 3. Student C correctly identifies this key insight. Choice A incorrectly assumes $a_1 = 21$ . Choice B uses the wrong starting index. Choice D arbitrarily sets $a_1 = 0$ without recognizing that any first term works.
An arithmetic sequence is defined by $a_1 = 20$ and $a_n = a_{n-1} + d$ for $n \geq 2$ , where $d < 0$ . If the sum $a_3 + a_7 = 24$ , what is the value of $a_5$ ?
Answer: $a_5 = 12$ When you encounter an arithmetic sequence problem, remember that each term follows the pattern $a_n = a_1 + (n-1)d$ , where $d$ is the common difference between consecutive terms. Let's work with what we know. Since $a_1 = 20$ and the sequence has common difference $d$ , we can express any term using the general formula. Specifically: - $a_3 = a_1 + 2d = 20 + 2d$ - $a_7 = a_1 + 6d = 20 + 6d$ Given that $a_3 + a_7 = 24$ , we can substitute: $(20 + 2d) + (20 + 6d) = 24$ $40 + 8d = 24$ $8d = -16$ $d = -2$ Now we can find $a_5 = a_1 + 4d = 20 + 4(-2) = 20 - 8 = 12$ . Looking at the wrong answers: Choice A ( $a_5 = 4$ ) would require $d = -4$ , which doesn't satisfy our constraint. Choice B ( $a_5 = 8$ ) corresponds to $d = -3$ , again inconsistent with $a_3 + a_7 = 24$ . Choice C ( $a_5 = 16$ ) implies $d = -1$ , which would make $a_3 + a_7 = 34$ , not 24. The correct answer is D: $a_5 = 12$ . Study tip: In arithmetic sequence problems, always use the constraint equations to find the common difference first. Once you have $d$ , you can find any term in the sequence. Also remember that the middle term of an arithmetic sequence equals the average of terms equidistant from it.
Consider the recursive rule $s_1 = 3$ and $s_n = s_{n-1} - 2.5$ for $n \geq 2$ . What is the smallest positive integer $k$ such that $s_k < 0$ ?
Answer: $k = 3$ Computing terms: $s_1 = 3$ , $s_2 = 3 - 2.5 = 0.5 > 0$ , $s_3 = 0.5 - 2.5 = -2 < 0$ . Therefore $k = 3$ is the smallest positive integer where $s_k < 0$ . Choice A gives the last positive term's index minus 1. Choice C assumes the pattern continues one more step. Choice D miscounts by assuming it takes 5 steps total.
The recursive rule $v_1 = 20$ and $v_n = v_{n-1} + k$ generates an arithmetic sequence. If the sum $v_1 + v_2 + v_3 = 66$ , what is the recursive rule for this sequence?
Answer: $v_1 = 20$ , $v_n = v_{n-1} + 2$ for $n \geq 2$ When you encounter recursive sequences with a constant difference, you're working with arithmetic sequences where each term is found by adding the same value $k$ to the previous term. To find $k$ , start by expressing the first three terms using the recursive rule. Given $v_1 = 20$ , you have: - $v_1 = 20$ - $v_2 = v_1 + k = 20 + k$ - $v_3 = v_2 + k = (20 + k) + k = 20 + 2k$ Since $v_1 + v_2 + v_3 = 66$ , substitute these expressions: $20 + (20 + k) + (20 + 2k) = 66$ $60 + 4k = 66$ $4k = 6$ $k = \frac{3}{2} = 1.5$ Wait—none of the answer choices show $k = 1.5$ . Let me recalculate more carefully. Actually, let me check option D by working backwards. If $k = 2$ : - $v_1 = 20$ - $v_2 = 20 + 2 = 22$ - $v_3 = 22 + 2 = 24$ - Sum: $20 + 22 + 24 = 66$ ✓ This confirms answer D is correct. Looking at the other options: Choice A gives $k = 46$ , which would make $v_2 = 66$ and $v_3 = 112$ , creating a sum far too large. Choice B gives $k = 6$ , resulting in terms 20, 26, 32 with sum 78. Choice C gives $k = 22$ , resulting in terms 20, 42, 64 with sum 126. Study tip: When solving recursive sequence problems, always verify your answer by calculating the actual terms and checking against given conditions—this catches calculation errors and confirms your solution.
A recursive rule is given as $a_1 = k$ and $a_n = a_{n-1} + 7$ for $n \geq 2$ . If $a_4 = 29$ , which equation can be used to find the value of $k$ ?
Answer: $k + 7 \cdot 3 = 29$ In an arithmetic sequence, aₙ = a₁ + (n-1)d. Here, a₄ = a₁ + (4-1)(7) = k + 21 = 29, so k + 7×3 = 29. Choice A incorrectly uses 4 instead of 3 steps. Choice C treats this as geometric growth rather than arithmetic. Choice D incorrectly multiplies k by 7 instead of adding multiples of 7.
An arithmetic sequence satisfies $u_3 = 14$ and $u_8 = 34$ . Which recursive rule correctly represents this sequence?
Answer: $u_1 = 6$ , $u_n = u_{n-1} + 4$ for $n \geq 2$ From $u_3 = 14$ to $u_8 = 34$ is 5 steps, so the common difference is $(34-14)/5 = 4$ . Working backwards: $u_2 = 14 - 4 = 10$ and $u_1 = 10 - 4 = 6$ . The recursive rule is $u_1 = 6$ , $u_n = u_{n-1} + 4$ for $n \geq 2$ . Choice B uses $u_3$ as the first term. Choice C uses the total difference instead of common difference. Choice D starts recursion at the wrong index.
A sequence follows the rule $b_1 = 12$ and $b_n = b_{n-1} - 4$ for $n \geq 2$ . Which statement best explains why this is classified as an arithmetic sequence?
Answer: The recursive formula subtracts the same value from each term, creating equal differences between consecutive terms An arithmetic sequence is defined by having a constant difference between consecutive terms. Since $b_n = b_{n-1} - 4$ , each term is exactly 4 less than the previous term, creating a constant difference of -4. Choice A focuses on the sign rather than the constant difference property. Choice B incorrectly states addition occurs. Choice D describes features of recursive sequences generally, not specifically arithmetic sequences.
An arithmetic sequence has the recursive rule $c_1 = m$ and $c_n = c_{n-1} - 5$ for $n \geq 2$ . If $c_6 = 12$ , and the sequence continues indefinitely, what is the smallest positive integer $k$ such that $c_k < 0$ ?
Answer: $k = 9$ First, find m: c₆ = m + (6-1)(-5) = m - 25 = 12, so m = 37. The sequence is: 37, 32, 27, 22, 17, 12, 7, 2, -3, ... We need the first term less than 0. c₈ = 37 - 7(5) = 2 > 0, and c₉ = 37 - 8(5) = -3 < 0. Therefore k = 9. Choice A gives c₈ = 2 > 0. Choice C gives c₇ = 7 > 0. Choice D gives c₁₀ = -8 < 0 but this isn't the smallest k.
The recursive rule $a_1 = k$ and $a_n = a_{n-1} + 5$ generates an arithmetic sequence where $a_6 = 23$ . What is the value of $k$ ?
Answer: $k = -2$ Working backwards from $a_6 = 23$ using the recursive rule $a_n = a_{n-1} + 5$ : $a_5 = 23 - 5 = 18$ , $a_4 = 18 - 5 = 13$ , $a_3 = 13 - 5 = 8$ , $a_2 = 8 - 5 = 3$ , $a_1 = 3 - 5 = -2$ . Therefore $k = -2$ . Choice B gives $a_2$ , choice C gives $a_3$ , and choice D gives $a_5$ .

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