Recursive rule

In the recursive rule, add the value of \(n\) for the current step to the previous term. Thus, \(a_2=2+1=3\), \(a_3=3+2=5\), and \(a_4=5+3=8\). Therefore, the correct answer is \(8\). \(9\) would result from incorrectly adding \(4\) in the final step. Exam tip: write the relevant value of \(n\) before calculating each next term.

Using the recursive rule, subtract 1, 2, 3, and then 4 successively: a_2=15-1=14, a_3=14-2=12, a_4=12-3=9, and a_5=9-4=5. Therefore, the correct answer is 5. The value 6 would result from incorrectly subtracting 3 instead of 4 in the final step. Exam tip: To find a_5, apply the rule for n=1 through n=4.

The recursive rule subtracts the current value of \(n\) to obtain the next term. Thus, \(a_2=15-1=14\), \(a_3=14-2=12\), and \(a_4=12-3=9\). Therefore, the correct answer is 9. Note that 12 is \(a_3\), not \(a_4\). Exam tip: to find \(a_4\), apply the rule successively for \(n=1,2,3\).

Starting with \(a_1=3\), apply the rule successively: \(a_2=3+2(1)=5\), \(a_3=5+2(2)=9\), \(a_4=9+2(3)=15\), and \(a_5=15+2(4)=23\). Hence, the correct answer is 23. The value 15 is \(a_4\), not \(a_5\). Exam tip: substitute \(n=1,2,3,\ldots\) one step at a time in a recursive rule.

The recursive rule adds consecutive odd numbers. Thus, \(a_2=6+1=7\), \(a_3=7+3=10\), \(a_4=10+5=15\), and \(a_5=15+7=22\). Hence, 22 is correct. The value 20 would result only if an increment were missed, so it is not correct. Exam tip: to find \(a_{n+1}\), use the value of \(n\) corresponding to the current term number.

The recursive rule makes each new term three times the preceding term. Thus, \(a_2=3\times4=12\) and \(a_3=3\times12=36\). Therefore, the correct answer is \(36\). \(48\) would result from multiplying 4 directly by 12, which does not follow the given rule. Exam tip: write \(a_2\) first before finding \(a_3\).

Apply the recursive rule step by step: \(a_2=2\times2+1=5\), \(a_3=2\times5+1=11\), and \(a_4=2\times11+1=23\). Therefore, the correct answer is 23. Note that 11 is \(a_3\), not \(a_4\). Exam tip: for each new term, first multiply the previous term by 2 and then add 1.

Apply the recursive rule one step at a time: \(a_2=2\times6-2=10\). Then \(a_3=2\times10-2=18\). Therefore, 18 is correct. The value 20 would result from forgetting to subtract 2 in the second step. Exam tip: calculate each term sequentially from the preceding term.

The recurrence rule says that every new term is obtained by adding 6 to the preceding term. The starting value is the first term, so each step increases the position by one and the value by 6. To reach the sixth term from the first term, there are five additions, not six. Consequently, the value is 1 plus 5 times 6, which equals 31. This is option D.

Writing the terms avoids an indexing mistake: a1 is 1, a2 is 7, a3 is 13, a4 is 19, a5 is 25, and a6 is 31. Equivalently, the arithmetic-sequence formula is a_n = a_1 + (n-1)d, with a_1 = 1, n = 6, and d = 6; hence a_6 = 1 + 5 times 6 = 31. The supplied answer D is correct. Values such as 25 correspond to the fifth term, so careful counting of the five transitions is essential.

The recursive rule subtracts 5 from each preceding term. Thus the sequence is 45, 40, 35, 30, 25, 20. Therefore, \(a_6=20\). The option 25 is \(a_5\), not \(a_6\). Exam tip: start counting from \(a_1\) and subtract 5 at each step.

The governing concept is a recursive rule in which each term, beginning with the third, is the sum of the two preceding terms. Start with a₁ = 1 and a₂ = 3. Then a₃ = a₂ + a₁ = 3 + 1 = 4. Next, a₄ = a₃ + a₂ = 4 + 3 = 7. Then a₅ = a₄ + a₃ = 7 + 4 = 11. Finally, a₆ = a₅ + a₄ = 11 + 7 = 18. Thus option C is correct. Option B is only the fifth term, and option A is not obtained at the sixth step. Option D would require an incorrect addition or an extra alteration of the recurrence. Writing every intermediate term prevents an index error.

The recursive rule says that each new term is the sum of the previous two terms. Thus, \(a_3=2+6=8\), \(a_4=6+8=14\), and \(a_5=8+14=22\). Therefore, 22 is correct. The value 14 is only \(a_4\), not \(a_5\). Exam tip: write every intermediate term in order until you reach the required term.

By the recursive rule, each new term is one-third of the preceding term. Thus, \(a_2=27\div3=9\), \(a_3=9\div3=3\), and \(a_4=3\div3=1\). Therefore, the correct answer is 1. Option 3 is \(a_3\), not \(a_4\). Exam tip: Count \(a_1\) as the first term.

By the recursive rule, each new term is 3 more than the previous term: 8, 11, 14, 17, 20. Therefore, 20 is the fifth term. The fourth term is 17, so it is not correct. Exam tip: list the terms along with their positions in recursive-sequence questions.

The governing concept is repeated addition in a recursive sequence. Beginning with a_1 = 6 and adding 6 each time produces 6, 12, 18, 24, 30. The number 30 appears after four additions, at position 5, so it is the fifth term. Therefore option B is correct. Position 4 contains 24, and positions 6 and 7 contain 36 and 42.

A recursive rule must state both the initial term and the operation that produces each following term. The sequence begins with 7, so a₁ must equal 7. Examine consecutive differences: 11 − 7 = 4, 15 − 11 = 4, and 19 − 15 = 4. The difference is consistently 4, which means each new term is obtained by adding 4 to the preceding term. Therefore the correct rule is a₁ = 7 and aₙ₊₁ = aₙ + 4, so option A is correct. Option B uses the wrong difference and would produce 7, 10, 13, 16. Option C incorrectly identifies 11 as the first term. Option D multiplies by 4, producing 7, 28, 112 rather than the given sequence.

A recursive description must state both the correct first term and the operation connecting consecutive terms. The sequence begins with 25, and each following term is 5 less: 25 − 5 = 20, 20 − 5 = 15, and 15 − 5 = 10. Hence option C is correct. Option A has the wrong first term, B increases instead of decreases, and D multiplies by 5.

A recursive rule defines each term from the term immediately before it, so we compare consecutive terms. The ratios are 12 ÷ 4 = 3, 36 ÷ 12 = 3, and 108 ÷ 36 = 3. Thus every new term is three times its predecessor. Since the first term is 4, the complete recursive description is a₁ = 4 and aₙ₊₁ = 3aₙ. Therefore option C is correct. Option A adds 8 and would produce 4, 12, 20, rather than the given sequence. Option B doubles each term and would produce 4, 8, 16. Option D uses the correct multiplier but starts with the wrong first term, so it cannot generate the stated sequence.

The governing concept is a recursive rule involving a fixed multiplier. The sequence starts at 100, and every term is half of the preceding term: 100 ÷ 2 = 50 and 50 ÷ 2 = 25. Thus the matching rule is a_1 = 100 and a_(n+1) = a_n/2, which is option C. A doubles, B has the wrong initial term, and D does not give 25 after 50.

The recurrence aₙ₊₁ = aₙ + 1 means that one is added whenever we move from one term to the next. Starting with a₁ = 12, calculate successively: a₂ = 12 + 1 = 13, a₃ = 13 + 1 = 14, a₄ = 14 + 1 = 15, and a₅ = 15 + 1 = 16. Therefore option C is correct. A shorter equivalent calculation is a₅ = a₁ + (5 − 1) × 1 = 12 + 4 = 16, because four transitions are needed to reach the fifth term. Option A stops too early, option B is the fourth term, and option D adds one extra increment.