Recursive rule

The governing concept is recursive subtraction: the amount removed changes with the index n, so the terms must be calculated successively. Start from a₁ = 150. For n = 1, subtract 5(1) + 4 = 9, giving a₂ = 150 - 9 = 141. For n = 2, subtract 5(2) + 4 = 14, giving a₃ = 141 - 14 = 127. For n = 3, subtract 5(3) + 4 = 19, so a₄ = 127 - 19 = 108. Equivalently, a₄ = 150 - (9 + 14 + 19) = 108. Therefore option A is correct. The other values may arise from adding instead of subtracting, using an incorrect index, or omitting one of the three required transitions.

Using the recursive rule with \(n=1\), \(a_2=150-(5\times1+4)=141\). Then, with \(n=2\), \(a_3=141-(5\times2+4)=141-14=127\). Therefore, \(127\) is correct. A close distractor such as \(124\) can result from calculating \(5n+4\) incorrectly in the second step. Exam tip: use the next value of \(n\) for each successive term.

(a_3=3\times2-2\times1+1=5) and (a_4=3\times5-2\times2+1=12). The correct option should include (12).

In an arithmetic progression, each term is obtained by adding a fixed common difference to the previous term. Since the difference is 3, \(a_n=a_{n-1}+3\). Exam tip: look for constant addition in a recursive AP rule.

The total addition is (\frac{3(1+2+3+4)}{2}=15), so (a_5=20). Exam tip: find the total fractional addition first.

The total subtraction is (\frac{3(1+2+3+4)}{2}=15), so (a_5=25). Exam tip: calculate total fractional subtraction first.

In the recursive rule, use the current index to find the next term. Thus, \(a_2=1\times2+3=5\), \(a_3=2\times5+3=13\), and \(a_4=3\times13+3=42\). Therefore, the correct answer is \(42\). An option such as \(39\) results from using the multiplier 3 or the added 3 incorrectly while finding \(a_4\). Exam tip: for \(a_{n+1}\), substitute \(n=1\), then \(n=2\), and then \(n=3\) in order.

Under \(a_{n+1}=r a_n\), the ratio \(a_{n+1}/a_n=r\) remains constant, so the sequence is geometric. In option B, the difference is constant, which defines an arithmetic progression. Exam tip: check whether a rule fixes a ratio or a difference.

The correct rule is \(a_1=3,\ a_{n+1}=2a_n+2\). It gives \(2\times3+2=8\), then \(2\times8+2=18\), and \(2\times18+2=38\), matching every listed term. Option B gives \(7\) as the second term, while option C has the wrong first term. Exam tip: test a recursive rule by generating at least three consecutive terms.

In option B, \(a_1=10\). Applying the rule gives \(2(10)-3=17\), \(2(17)-3=31\), and \(2(31)-3=59\), so it matches every given term. In option A, the next term would be 24, while option D has the wrong first term. Exam tip: when checking a recursive rule, verify the initial term and at least two subsequent terms.

The governing concept is a recursive sequence: each new term is calculated from the preceding term, and the value of n identifies the term currently being used. Start with a₁ = 5. For n = 1, a₂ = a₁ + 1² = 5 + 1 = 6. For n = 2, a₃ = 6 + 2² = 10. For n = 3, a₄ = 10 + 3² = 19. For n = 4, a₅ = 19 + 4² = 35. Therefore, 35 is the fifth term, so option B is correct. The common error is to square the index of the new term rather than the n appearing in the recurrence; that changes the sequence and may lead to another option.

The recursive rule subtracts 1², 2², 3², and so on. Thus, a₁=100, a₂=100−1=99, a₃=99−4=95, a₄=95−9=86, and a₅=86−16=70. Therefore, 70 is the fifth term. The fourth term is 86, so it is a close but incorrect option. Exam tip: in a recursive rule, use the value of n while moving from one term to the next.

Apply the recursive rule step by step: \(a_3=2+7+3=12\), \(a_4=7+12+4=23\), and \(a_5=12+23+5=40\). Therefore, the correct answer is 40. A value such as 38 can result from forgetting to add 5 while finding \(a_5\). Exam tip: for every new term, add both preceding terms and the index of the term being found.

First find the third term using the recursive rule: (a_3=7+2+3=12). Then, (a_4=a_3+a_2+4=12+7+4=23). Therefore, the correct answer is 23. The value 21 may result from forgetting to add 4 in the final step, but the rule requires adding the current index n. Exam tip: apply a recursive rule step by step until you reach the required term.

Given \(a_1=9\), first use \(n=1\) in the recursive rule: \(a_2=a_1+a_1+1^2=9+9+1=19\). Next, use \(n=2\): \(a_3=a_2+a_1+2^2=19+9+4=32\). Hence, the correct answer is \(32\). A choice such as \(34\) can result from an error in using the square term or the previous term. Exam tip: substitute the correct value of \(n\) at every step of a recurrence relation.

Given \(a_1=9\), substitute \(n=1\) in the recursive rule to find the second term: \(a_2=a_1+a_1+1^2=9+9+1=19\). Hence, 19 is correct. The value 17 would result from incorrectly omitting the \(n^2\) term. Exam tip: To find \(a_{n+1}\), first substitute the appropriate value of \(n\) into the recursive rule.

The recursive rule generates each new term from the preceding term. Thus, \(a_2=2\times6+1=13\), \(a_3=2\times13+2=28\), and \(a_4=2\times28+3=59\). Hence, the correct answer is 59. An answer such as 57 can result from incorrectly adding 1 instead of using \(n=3\) in the final step. Exam tip: write the value of \(n\) for each transition before substituting it into the rule.

The governing concept is a recursive sequence: each term is calculated from the preceding term using the current value of n. Begin with a₁ = 80. For n = 1, subtract 3(1)+2 = 5, so a₂ = 80 - 5 = 75. For n = 2, subtract 3(2)+2 = 8, giving a₃ = 75 - 8 = 67. For n = 3, subtract 3(3)+2 = 11, so a₄ = 67 - 11 = 56. Finally, for n = 4, subtract 3(4)+2 = 14, giving a₅ = 56 - 14 = 42. Thus option B is correct. The successive deductions are 5, 8, 11, and 14; each rises by 3. Skipping a transition or using the wrong index can produce the other choices.

Substituting \(n=1,2,3,4\) into the recursive rule gives successive increments of \(4,10,18,28\). Thus, \(a_2=1+4=5\), \(a_3=5+10=15\), \(a_4=15+18=33\), and \(a_5=33+28=61\). Therefore, 61 is correct. A value such as 63 would result from an incorrect addition at one of the steps. Exam tip: to find \(a_5\), apply the rule from \(n=1\) through \(n=4\).

In an arithmetic progression, the difference between consecutive terms is constant. The rule \(a_{n+1}=a_n+7\) adds exactly 7 at every step, so its common difference is 7. In \(a_n+n\), the difference changes with \(n\). Exam tip: compute \(a_{n+1}-a_n\); it must be a constant for an AP.