Recursive rule

By the recursive rule, each new term is the sum of the two preceding terms. Thus, \(a_3=4+1=5\) and \(a_4=5+4=9\). Therefore, the correct answer is 9. A value such as 10 results from adding incorrect terms. Exam tip: find \(a_3\) first, then use it to calculate \(a_4\).

The recursive rule divides each term by 3 to obtain the next term. Thus, \(a_2=\frac{12}{3}=4\), and \(a_3=\frac{4}{3}\). Option 2 would result from dividing 12 by 3 only once, so it is \(a_2\), not \(a_3\). Exam tip: to find \(a_3\) from \(a_1\), apply the recursive rule twice.

The recursive rule increases each term by 2. Thus, the terms are 3, 5, 7, 9, 11. Therefore, 11 is the fifth term. The sixth term would be 13, so the sixth option is not correct. Exam tip: list the terms in order and count their positions.

The governing concept is a recursive rule with a constant increase of 4. Starting from a_1 = 4, the terms are 4, 8, 12, 16, and 20. Counting from the first position shows that 20 is a_5, the fifth term. Therefore option B is correct. The fourth term is only 16, while the sixth and seventh terms would be 24 and 28.

A recursive rule must contain two parts: an initial value and a relationship that generates each next term from the preceding term. The first term here is 5. The successive differences are 8−5=3, 11−8=3, and 14−11=3, so the same operation is repeated each time: add 3 to the previous term. Consequently, the correct rule is a_1 = 5 and a_(n+1) = a_n + 3, which is option A. Option B uses a difference of 4 and would produce 5, 9, 13, ... . Option C starts with the second term rather than the first, and option D doubles each term, producing a completely different sequence. Both the starting value and recurrence must agree with the displayed sequence.

A recursive rule describes a sequence by giving its starting term and a rule for obtaining each next term from the preceding term. In the sequence \(20,15,10,5,\ldots\), the first term is clearly 20. Comparing consecutive terms shows that each term is 5 less than the term immediately before it.

Starting with \(a_1=20\), the rule \(a_{n+1}=a_n-5\) produces \(a_2=15\), then \(a_3=10\), and then \(a_4=5\), exactly matching the sequence. Adding 5 would make the sequence increase, and starting with 15 would shift every position. Multiplication by 5 also does not fit. Therefore option C is correct.

A recursive rule must state the initial term and explain how to obtain each following term from the preceding one. The sequence starts with a_1=2. Each next value is twice the previous value: 4=2×2, 8=2×4, and 16=2×8. Therefore the recursive rule is a_1=2 and a_(n+1)=2a_n, which is option B. Option A adds 2 and would produce 2, 4, 6, 8 instead. Option C has the correct multiplier but starts with 4, so it describes a different indexed sequence. Option D subtracts 2 and produces decreasing terms. The constant multiplier also shows the related geometric pattern.

A recursive rule requires both the correct initial term and the correct transition from one term to the next. The sequence begins with a_1=81. Dividing successively by 3 gives 81÷3=27, 27÷3=9, and 9÷3=3. Thus the rule is a_1=81 and a_(n+1)=a_n÷3, which is option C. Option A multiplies by 3 and would produce 243 after 81. Option B uses division by 3 but starts at 27, so it does not match the given first term or indexing. Option D subtracts 3 and would produce 78 after 81. The repeated quotient also identifies the sequence as geometric with common ratio 1/3.

The governing concept is a recursively defined sequence: each new term is obtained from the preceding term by applying the stated rule. Start with a₁ = 7 and add 1 repeatedly until the sixth position is reached. Thus a₂ = 7 + 1 = 8, a₃ = 9, a₄ = 10, a₅ = 11, and a₆ = 12. Therefore option C is correct. Equivalently, this is an arithmetic sequence with first term 7 and common difference 1, so aₙ = 7 + (n − 1)×1; putting n = 6 gives 7 + 5 = 12. Option A stops at a₄, option B is a₅, and option D applies the rule one extra time to obtain a₇.

The recursive rule makes each new term 10 less than the previous term. Thus, a_2=40, a_3=30, and a_4=20. Therefore, 20 is correct. The value 30 is the third term, not the fourth. Exam tip: for a recursive sequence, list the terms step by step up to the required term.

The recursive rule increases each term by 2. Thus the terms are 6, 8, 10, 12, 14. Hence \(a_3=10\) and \(a_5=14\), so \(a_5-a_3=14-10=4\). The value 2 is the increase for one step only; there are two steps from \(a_3\) to \(a_5\). Exam tip: when subtracting two terms, count the number of steps between their indices.

The recursive rule increases each term by 4. Thus, \(a_2=3+4=7\), \(a_3=11\), and \(a_4=15\). Therefore, \(a_2+a_4=7+15=22\), so option C is correct. \(20\) can result from an incorrect use of the common increase. Exam tip: write the required terms in order before finding their sum.

Given a_1=2, the recursive rule gives a_2=a_1^2=2^2=4. Then a_3=a_2^2=4^2=16. Hence, 16 is correct. The value 8 is the cube of 2, but this rule requires squaring each preceding term. Exam tip: in a recursive sequence, apply the rule step by step until the required term is reached.

Given \(a_1=1\), \(a_2=a_1^2+1=1^2+1=2\). Then \(a_3=a_2^2+1=2^2+1=5\). Therefore, the correct answer is 5. Option 4 is only \(2^2\); the recursive rule also requires adding 1. Exam tip: for every new term, square the preceding term and then add 1.

The recursive rule subtracts 15 from each preceding term. Thus, \(a_2=100-15=85\), and then \(a_3=85-15=70\). Option 85 is a close distractor because it is the second term, not the third term. Exam tip: write each step in a recursive sequence until you reach the required term.

The recursive rule says to add 6 to the previous term each time. Thus, a_2=15, a_3=21, and a_4=27. The value 21 is the third term, so it is a close but incorrect option. Exam tip: from a_1 to a_4, apply the rule three times.

The recursive rule makes each new term four times the preceding term. Thus, a_2=4×2=8 and a_3=4×8=32. Therefore, 32 is correct. The value 64 would be a_4, not a_3. Exam tip: to find a_3 from a_1, apply the rule twice.

In option A, each new term is formed by adding 6 to the previous term, so \(a_{n+1}-a_n=6\) is constant. This is the defining property of an AP. In option D, the difference changes with \(n\). Exam tip: check for a constant difference.

Put \(n=1\) in the recursive rule: \(a_2=\frac{a_1}{2}+1\). Therefore, \(a_2=\frac{40}{2}+1=20+1=21\). Option 20 only halves 40; the rule also requires adding 1. Exam tip: to find \(a_{n+1}\), substitute the given term first and then follow the order of operations.

In a recursive rule, the value of n changes for each new term. Thus, a_2=1+3(1)=4, a_3=4+3(2)=10, and a_4=10+3(3)=19. Therefore, the correct answer is 19. The value 16 would result from incorrectly adding only 6 in the last step. Exam tip: To find a_4, apply the rule successively for n=1, 2, and 3.