Recursive rule

A recursive rule generates each term from the preceding one, so it needs \(a_1\) to begin. Then \(a_2=2a_1+5\). A common difference is relevant only for an arithmetic progression. Exam tip: always identify the starting value.

Each new term is obtained by adding the triangular quantity n(n+1)/2, where n is the index of the current term. To find a5, begin with a1 and perform the recurrence for n = 1, 2, 3, and 4. The added quantities are 1, 3, 6, and 10. Starting from 3, the terms become 4, 7, 13, and 23. Therefore a5 = 23, which is option B.

The calculation is a careful index exercise: a2 = 3 + 1(2)/2 = 4; a3 = 4 + 2(3)/2 = 7; a4 = 7 + 3(4)/2 = 13; and a5 = 13 + 4(5)/2 = 23. The increments are the triangular numbers 1, 3, 6, and 10, but they must be attached to the correct transition. The supplied answer B and explanation are consistent with the recurrence and contain no mathematical error.

Using \(n=1,2,3\), the successive amounts subtracted are \(1,3,6\). Thus, \(a_2=70-1=69\), \(a_3=69-3=66\), and \(a_4=66-6=60\). Therefore, the correct answer is 60. The value 58 would result from incorrectly taking the final decrement as 8 instead of 6. Exam tip: to find \(a_4\) from \(a_1\), apply the recursive rule three times.

Use the recursive rule first with \(n=1\): \(a_2=\frac{10}{2}+2(1)=7\). Then use \(n=2\): \(a_3=\frac{7}{2}+2(2)=\frac{15}{2}\). The value \(7\) is \(a_2\), not \(a_3\). Exam tip: while finding successive terms, substitute the correct index value of \(n\) each time.

To find \(a_2\), substitute \(n=1\) and \(a_1=12\) in the recursive rule: \(a_2=\frac{a_1}{2}+2(1)=\frac{12}{2}+2=8\). Hence, 8 is correct. The value 6 is only half of 12 and misses the \(2n\) term. Exam tip: when finding \(a_{n+1}\), first use the correct value of \(n\), then substitute the corresponding term \(a_n\).

Given \(a_1=6\), we have \(2a_1=12\). First, put \(n=1\): \(a_2=a_1+2a_1+1=6+12+1=19\). Next, put \(n=2\): \(a_3=a_2+2a_1+2=19+12+2=33\). Therefore, the correct answer is 33. The value 31 can result from forgetting to add \(n=2\) in the second step. Exam tip: in a recursive rule, substitute the correct value of \(n\) for each successive term.

Each new term is obtained by adding 5, so \(a_{n+1}-a_n=5\) is constant. Hence it is an arithmetic progression. A geometric progression requires a constant ratio. Exam tip: check consecutive differences first.

Using the recursive rule, first find the third term and then the fourth term. \(a_3=a_2+a_1+3=4+2+3=9\). Therefore, \(a_4=a_3+a_2+4=9+4+4=17\). Hence, the correct answer is 17. A value such as 16 can result from adding the \(+n\) term incorrectly. Exam tip: always substitute the current value of \(n\) separately for each new term.

Substitute \(n=3\) into the recursive rule: \(a_3=a_2+a_1+3=4+2+3=9\). Therefore, the correct answer is \(9\). Adding only \(a_2\) and \(a_1\) gives \(6\), but the extra index term \(3\) must also be included. Exam tip: always substitute the correct value of \(n\) when using a recursive rule.

By the recursive rule, each new term is obtained by squaring the previous term and subtracting that same previous term. Thus, \(a_2=3^2-3=6\) and \(a_3=6^2-6=30\). Therefore, the correct answer is 30. The value 36 is only \(6^2\); the rule also requires subtracting 6. Exam tip: always find \(a_2\) before calculating \(a_3\).

In a recursive rule, each new term is calculated from the preceding term. For


a_1=2,


a_2=a_1^2+a_1=2^2+2=6.

Then,


a_3=a_2^2+a_2=6^2+6=36+6=42.

Therefore, the correct answer is 42. The close distractor 36 is only


a_2^2=6^2

and misses the required addition of


a_2=6

in the rule. Exam tip: at every step, square the previous term and then add that same term.

The recursive rule increases each term by 3. To reach the eighth term, add 3 seven times to the first term: \(a_8=4+7\times3=25\). Therefore, 25 is correct. The value 22 is the seventh term, so it is a close but incorrect distractor. Exam tip: from \(a_1\) to \(a_n\), the number of increases is \(n-1\).

This is an arithmetic sequence because every term is obtained by adding the same number, 7, to the previous term. Starting at 9, the terms are 9, 16, 23, 30, 37, 44, and 51. The value 51 therefore appears as the seventh term, so option B is correct. The important point is that the first term already has index 1; it is not counted as a zero-th term.

Using the arithmetic-sequence formula gives the same result: a_n = a_1 + (n-1)d. Here, 51 = 9 + (n-1)7, so 42 = 7(n-1), which gives n-1 = 6 and n = 7. Listing the terms is often safer for a short question and confirms the result directly. The supplied answer B and its explanation correctly identify the seventh position. Option A would be one step too early and gives 44, not 51.

Using the recursive rule, \(a_2=2(4)+1=9\), \(a_3=2(9)+1=19\), and \(a_4=2(19)+1=39\). Therefore, \(a_4-a_2=39-9=30\). Option 28 may result from an error while generating the terms or subtracting them. Exam tip: apply a recursive rule step by step to each preceding term until the required term is reached.

Using the recursive rule, \(a_3=6+3+2=11\), \(a_4=11+6+2=19\), and \(a_5=19+11+2=32\). Therefore, the correct answer is 32. An answer such as 31 may result from adding the constant term 2 incorrectly. Exam tip: while finding each new term, include both previous terms and the given constant term.

The recursive rule forms each new term by adding the previous two terms and 2. Thus, \(a_3=6+3+2=11\), and \(a_4=11+6+2=19\). Therefore, the correct answer is 19. A value such as 17 can result from forgetting to add 2 at one step. Exam tip: calculate \(a_3\) first before finding \(a_4\).

Since 7 is added to each preceding term, the difference between consecutive terms is constantly 7. Hence it is an AP, not a GP. Exam tip: check \(a_{n+1}-a_n\) for a constant difference.

In a recursive rule, the value of n changes at every step. For n=1, \(a_2=2+5(1)-1=6\). For n=2, \(a_3=6+5(2)-1=15\). Then, for n=3, \(a_4=15+5(3)-1=29\). Therefore, the correct answer is 29. A choice such as 27 can result from using an incorrect increment instead of \(14\) in the final step. Exam tip: to find \(a_4\), substitute \(n=1,2,3\) successively.

Apply the recursive rule successively for \(n=1,2,3\): \(a_2=120-(3\times1+2)=115\), \(a_3=115-(3\times2+2)=107\), and \(a_4=107-(3\times3+2)=96\). Hence, the correct answer is \(96\). A value such as \(99\) can result from using an incorrect value of \(n\) in the amount being subtracted. Exam tip: to find \(a_4\) from \(a_1\), apply the recursive rule three times.

Apply the recursive rule successively for n=1, 2, and 3. Thus, \(a_2=2+1^2+1+1=5\), \(a_3=5+2^2+2+1=12\), and \(a_4=12+3^2+3+1=25\). Therefore, the correct answer is 25. A value such as 23 can result from calculating the final increment \(3^2+3+1=13\) incorrectly. Exam tip: To find \(a_4\), apply the rule three times.

Apply the recursive rule first with \(n=1\): \(a_2=a_1+1^2+1+1=2+3=5\). Next, with \(n=2\), \(a_3=a_2+2^2+2+1=5+7=12\). Therefore, the correct answer is 12. The value 14 would result from an incorrect calculation of the second increase. Exam tip: change the value of \(n\) at every recursive step.

The recurrence can be rewritten as \(a_{n+1}=a_n+\frac{a_n}{4}=\frac{5}{4}a_n\). Thus every new term is \(\frac{5}{4}\) times the previous term. Beginning with \(a_1=16\), the second term is \(a_2=16+\frac{16}{4}=16+4=20\). Applying the same rule again gives \(a_3=20+\frac{20}{4}=20+5=25\). Therefore option C is correct.

It is important to use the newly obtained term when calculating the next one. The second step uses 20, not the original 16. This is why the added amount changes from 4 to 5. The sequence begins \(16,20,25\), so the required third term is 25. The rule is a multiplicative growth rule with common factor \(\frac{5}{4}\), although the recursive form is the one asked for.

Apply the recursive rule step by step: \(a_3=3(2)-2(1)=4\), \(a_4=3(4)-2(2)=8\), and \(a_5=3(8)-2(4)=16\). Therefore, the correct answer is 16. A value such as 15 may result from using an incorrect coefficient or subtraction. Exam tip: in a second-order recurrence, write down both preceding terms before calculating the next term.

(a_3=3\times2-2\times1=4) and (a_4=3\times4-2\times2=8). Exam tip: use both previous terms in the rule.

Using the recursive rule successively for n=1, 2, 3, and 4: a_2=9+2(1)-2=9, a_3=9+2(2)-2=11, a_4=11+2(3)-2=15, and a_5=15+2(4)-2=21. Therefore, the correct answer is 21. A value such as 23 can result from applying the increment 2n-2 incorrectly. Exam tip: while finding each next term, use n equal to the position of the preceding term.