Sequences

This is a geometric sequence in which each term is obtained by multiplying the previous term by 3: 9, 27, 81, 243, 729. Therefore, 729 is the 5th term. The 4th term is 243, so it is not correct. Exam tip: List the terms in order and check for a common multiplier.

This is an arithmetic sequence with first term 15 and common difference 9. Therefore, the nth term is \(a_n=15+(n-1)\times 9\). Putting \(a_n=105\), \(15+(n-1)\times9=105\), so \(n-1=10\) and \(n=11\). Hence, 105 is the 11th term. The 10th term is 96, so it is not correct. Exam tip: use \(a_n=a+(n-1)d\) to find the position of a term in an arithmetic sequence.

This is an arithmetic progression with first term 170 and common difference \(-12\). Therefore, \(a_n=170+(n-1)(-12)\). Setting \(a_n=74\) gives \(170-12(n-1)=74\), so \(n-1=8\) and \(n=9\). Hence, 74 is the 9th term. The 8th term is 86, so it is not correct. Exam tip: To find the position of a term, substitute its value in \(a_n=a+(n-1)d\).

The differences between consecutive terms are \(20-9=31-20=42-31=11\), so this is an arithmetic progression. Its nth term is \(a_n=a_1+(n-1)d=9+(n-1)\times11=11n-2\). In option B, putting \(n=1\) gives 20 as the first term, so it is incorrect. Exam tip: To check a linear rule quickly, substitute \(n=1\) and then \(n=2\).

This is an arithmetic sequence because each successive term increases by 9. Here, \(a_1=16\) and \(d=9\). Therefore, \(a_n=a_1+(n-1)d=16+9(n-1)=9n+7\), so option A is correct. In option B, substituting \(n=1\) gives 25 as the first term, not 16. Exam tip: Check a proposed \(a_n\) rule by putting \(n=1\) to verify the first term.

Substituting \(n=1,2,3,4\) in the rule gives \(a_1=8(1)-5=3\), \(a_2=11\), \(a_3=19\), and \(a_4=27\). Therefore, the first four terms are \((3,11,19,27)\). The option \((8,16,24,32)\) uses only \(8n\) and ignores the subtraction of 5. Exam tip: start with \(n=1\) to find the first term.

Substitute n=6: a_6=3(6^2)+4(6)=3(36)+24=108+24=132. Therefore, the correct answer is 132. An option such as 128 can result from an error in squaring or multiplication. Exam tip: evaluate powers first, then multiply, and finally add.

To find the eighth term, substitute n=8 in the rule: \(a_8=140-9(8)=140-72=68\). Hence, 68 is correct. 66 may result from an incorrect subtraction of \(140-72\), while 72 is only the value of \(9\times8\), not the term itself. Exam tip: for an nth-term rule, multiply first and then subtract.

Given \(a_n=7n^2\), substitute \(n=1,2,3,4\). This gives \(7\times1^2=7\), \(7\times2^2=28\), \(7\times3^2=63\), and \(7\times4^2=112\). Hence, the sequence is \(7, 28, 63, 112\). Option A follows \(7n\), not \(7n^2\). Exam tip: For a rule-based sequence, substitute the first few natural-number values of \(n\).

For the first four terms, take n=1,2,3,4. Thus, a_1=3^1+1=4, a_2=3^2+2=11, a_3=3^3+3=30, and a_4=3^4+4=85. Therefore, the correct sequence is (4,11,30,85). Option B contains only powers of 3 and does not add n. Exam tip: evaluate 3^n first and then add n for each term.

Both sequences have common difference (7), so the term difference always remains (4). The (12)th terms also differ by (4).

Both sequences have difference (8), and their first terms differ by (9). So their (9)th terms also differ by (9).

This is an arithmetic progression with first term \(a=8\), common difference \(d=8\), and \(n=9\) terms. Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), we get \(S_9=\frac{9}{2}[16+8\times8]=\frac{9}{2}\times80=360\). Hence, 360 is correct. The value 384 may result from using an incorrect number of terms or last term. Exam tip: identify \(a\), \(d\), and \(n\) before applying the sum formula.

Each successive term decreases by 11. Therefore, the first 6 terms are 132, 121, 110, 99, 88, and 77. Their sum is 132 + 121 + 110 + 99 + 88 + 77 = 627. Hence, 627 is the correct option. The value 626 is one less than the required sum. Exam tip: In a decreasing sequence, first identify the common difference, list the required terms, and then add them.

This is an arithmetic progression with first term 10 and common difference 8. Hence, \(a_n=10+(n-1)\times8\). Putting \(a_n=90\), we get \(10+8(n-1)=90\), so \(n-1=10\) and \(n=11\). Therefore, 90 is the 11th term. The 10th term is \(82\), so the closest distractor is not correct. Exam tip: To find the position of a given term, substitute it in \(a_n=a+(n-1)d\).

This is an arithmetic progression with first term 11 and common difference 6. Hence, \(a_8=11+7\times6=53\) and \(a_9=11+8\times6=59\). Therefore, \(a_8+a_9=53+59=112\), so option D is correct. Although 110 is close, it is not the correct sum. Exam tip: use \(a_n=a+(n-1)d\) to find the \(n\)th term.

In sequence B, \(2=1+1\), \(3=1+2\), and \(5=2+3\), so every new term is the sum of the previous two. Sequence A has a constant difference of 2, not this sum rule. Exam tip: test the rule using any three consecutive terms.

Check consecutive differences: −7−(−11)=4 and −3−(−7)=4. As the difference remains constant, this is an AP. Square numbers do not have a constant difference. Exam tip: compare at least two differences.

In option C, the consecutive differences are \(3,6,12\), which are not equal, so it is not an AP. Its terms are multiplied by 2 each time. Exam tip: compare consecutive differences first.

In A, \((7+(-1))/2=3\) and the common difference is \(-4\), so every middle term is the mean of its neighbours. In B, differences \(4\) and \(8\) change. Exam tip: compare successive differences.