Recursive rule

Using the recursive rule, first put n=1: a_2=180-(6×1+5)=169. Next, put n=2: a_3=169-(6×2+5)=169-17=152. Therefore, 152 is correct. The value 150 would result from incorrectly subtracting 19 in the second step. Exam tip: remember that the value of n changes to 1, 2, 3, and so on for successive terms.

First, use the recursive rule to get \(a_3=4a_2-3a_1+2=4(4)-3(2)+2=12\). Then \(a_4=4a_3-3a_2+2=4(12)-3(4)+2=38\). Therefore, 38 is the correct answer. A value such as 36 can result from forgetting the constant term \(+2\). Exam tip: write the two preceding terms separately before applying their coefficients.

In \(a_n=a_{n-1}+a_{n-2}\), every new term depends on the two preceding terms, so both \(a_1\) and \(a_2\) are needed. Exam tip: count the previous terms used to identify the required initial terms.

The total addition is (\frac{5(1+2+3+4)}{2}=25), so (a_5=31). Exam tip: find the total fractional addition first.

The recurrence must be applied four times because the sequence moves from a₁ to a₅ through four transitions. For n = 1, 2, 3, and 4, the amounts subtracted are 5/2, 5, 15/2, and 10. Their total is (5/2)(1 + 2 + 3 + 4) = (5/2)(10) = 25. Consequently, a₅ = 60 − 25 = 35, so option C is correct. A term-by-term check gives a₂ = 57.5, a₃ = 52.5, a₄ = 45, and a₅ = 35. A common mistake is to apply the recurrence only three times or to read 5n/2 as 5n divided incorrectly. The total-subtraction method confirms the result.

In a recursive rule, substitute the current index value at each step. For \(n=1\), \(a_2=1\times3+4=7\). Then, for \(n=2\), \(a_3=2\times7+4=18\), and for \(n=3\), \(a_4=3\times18+4=58\). Therefore, 58 is correct. An answer such as 52 may result from not changing \(n\) correctly at every step. Exam tip: write down the value of \(n\) before calculating each next term.

In an AP, the difference between consecutive terms is constant. In A, \(a_{n+1}-a_n=-5\) for every \(n\), so the common difference is \(-5\). In C, the difference \(-n\) changes. Exam tip: first check \(a_{n+1}-a_n\).

The governing concept is a recursive rule, which specifies an initial term and an operation for obtaining each following term from the preceding term. To verify a rule, start with the stated first term and apply the recurrence repeatedly; it must reproduce all displayed terms. For option A, a₁ = 5. Then a₂ = 3(5) + 1 = 16, a₃ = 3(16) + 1 = 49, and a₄ = 3(49) + 1 = 148. Thus it matches the entire sequence. Option B produces 16 first but then 2(16) + 6 = 38, not 49. Option C begins with the wrong initial term, and option D gives 14 as the second term. Therefore A is the only correct rule.

A correct recursive rule must reproduce the starting term and every displayed transition. The successive decreases are 20−15 = 5, 15−7 = 8, and 7−(−8) = 15? Check the stated pattern carefully: using 3n+2 gives 5 for n=1, 8 for n=2, and 11 for n=3, which would produce −4, not −8. Therefore the supplied sequence and options are inconsistent; no option is unambiguously correct without correcting the fourth term to −4.

This is a recursive sequence because each new term uses the preceding term and the current index n. Starting with a₁ = 7, calculate successively: for n = 1, a₂ = 7 + 1² + 1 = 9; for n = 2, a₃ = 9 + 2² + 2 = 15; for n = 3, a₄ = 15 + 3² + 3 = 27; and for n = 4, a₅ = 27 + 4² + 4 = 47. Therefore 47 is the fifth term, making option B correct. The fourth term is only 27, while the sixth and seventh terms require additional recurrence steps. The essential distinction is between a term’s value, 47, and its position, 5.

The terms are (150,148,142,130,110), so (110) is the fifth term. Exam tip: write decreasing terms with positions.

Apply the recursive rule using the two preceding terms. \(a_3=8+2(2)+3=15\), \(a_4=15+2(8)+4=35\), and \(a_5=35+2(15)+5=70\). Therefore, the correct answer is 70. A value such as 66 can result from an error in calculating the \(2a_{n-2}\) term. Exam tip: write the current value of \(n\) separately at every step.

The rule \(a_n=3a_{n-1}-2a_{n-2}\) depends on the two preceding terms, so both \(a_1\) and \(a_2\) are needed. A and B are first-order rules and need only one initial term. Exam tip: count the previous terms used in the rule.

In the recursive rule, the fixed first term \(a_1=6\) and the relevant \(n^2\) are added at each step. Thus, \(a_2=6+6+1^2=13\), \(a_3=13+6+2^2=23\), and \(a_4=23+6+3^2=38\). Therefore, the correct answer is \(38\). A close option such as \(37\) may result from evaluating \(3^2\) incorrectly. Exam tip: to find \(a_4\), use \(n=3\) in the final recurrence step, not \(n=4\).

Use the recursive rule step by step. For \(n=1\), \(a_2=a_1+a_1+1^2=6+6+1=13\). Then, for \(n=2\), \(a_3=a_2+a_1+2^2=13+6+4=23\). Therefore, the correct answer is 23. The value of \(a_1\) remains 6 in every step; replacing it with \(a_2\) would be incorrect. Exam tip: To find \(a_3\), apply the rule first with \(n=1\) and then with \(n=2\).

The added values (2!,3!,4!) are (2,6,24), so (a_4=37). Exam tip: calculate factorial increments in order.

In a recursive rule, each new term is calculated from the preceding term. \(a_2=3(4)+1=13\), \(a_3=3(13)+2=41\), and \(a_4=3(41)+3=126\). Therefore, the correct answer is 126. A value such as 122 can result from forgetting to add \(+3\) in the final step. Exam tip: at each step, use the value of \(n\) from the subscript of the preceding term.

The rule subtracts a different amount at each step. Since the first term is \(a_1=150\), use \(n=1,2,3,4\) to reach \(a_5\). The subtracted quantities are found from \(n^2+2n\): for these four indices they are 3, 8, 15, and 24. Writing them separately reduces the chance of using the wrong index or missing a subtraction.

Apply the rule successively: \(a_2=150-3=147\), \(a_3=147-8=139\), \(a_4=139-15=124\), and \(a_5=124-24=100\). The same result comes from \(150-(3+8+15+24)=150-50=100\). Therefore option B is correct. The answer is not obtained by subtracting the same amount each time.

This is a second-order recursive sequence because each new term depends on the two immediately preceding terms. The recurrence is aₙ = 2aₙ₋₁ + 3aₙ₋₂, so the previous term is multiplied by 2 and the term before it by 3. Starting with a₁ = 2 and a₂ = 5, calculate a₃ = 2(5) + 3(2) = 16. Next, a₄ = 2(16) + 3(5) = 32 + 15 = 47. Finally, a₅ = 2(47) + 3(16) = 94 + 48 = 142. Hence option B is correct. The other choices can result from omitting one preceding term, interchanging coefficients, or making an arithmetic error. Both initial values are essential because this recurrence has order two.

First, substitute \(n=3\) in the recurrence: \(a_3=3\times3-1+3=11\). Then, for \(n=4\), \(a_4=3\times11-3+4=34\). Therefore, the correct answer is 34. The distractor 33 results from omitting the final \(+4\). Exam tip: write the current value of \(n\) separately at each step before substituting.