Sequences

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.

The recursive rule is to multiply the preceding term by 3 and then add 2. Therefore, the term after 161 is \(3\times161+2=483+2=485\). Option 483 is only three times 161; it misses the final addition of 2. Exam tip: apply each stated operation to the last given term in order.

Using the rule, first divide the previous term \(18.25\) by \(2\), then subtract \(1\): \(18.25\div 2-1=9.125-1=8.125\). Therefore, the next term is \(8.125\). \(9.125\) is only the quotient before subtracting \(1\). Exam tip: In recursive sequences, follow the stated order of operations carefully.

The successive differences are \(8,16,24,32\). Each difference increases by \(8\), so the next difference is \(40\). Therefore, the next term is \(90+40=130\). Choosing \(126\) would incorrectly use \(36\) as the next difference. Exam tip: for such sequences, first list consecutive differences and check their pattern.

The differences between consecutive terms are subtractions of 15, 30, 45, and 60. The amount subtracted increases by 15 each time, so the next subtraction is 75. Therefore, the next term is \(210-75=135\). Option 150 would result from subtracting 60 again, but the subtraction pattern is increasing. Exam tip: In such sequences, first write the differences between consecutive terms and look for their pattern.

The sequence follows a multiplication pattern in which the factors increase by one each time. Starting with 5, multiplying by 2 gives 10; multiplying that by 3 gives 30; and multiplying by 4 gives 120. The next factor in the pattern is therefore 5. The required next term is 120 times 5, or 600, so the correct choice is option C. The pattern is based on successive multipliers, not on adding a fixed number.

The arithmetic can be verified directly. The ratios between neighboring terms are 10 divided by 5, which is 2; 30 divided by 10, which is 3; and 120 divided by 30, which is 4. Continuing with ratio 5 gives 120 times 5 equals 600. Thus option C is exactly the value predicted by the visible rule. A value such as 480 would use a factor of 4 again and would stop the increasing-factor pattern. The supplied answer and explanation are therefore accurate and complete for the intended interpretation.

The third term of the first sequence is 252, and that of the second sequence is 180. Therefore, the required ratio is 252:180. Dividing both terms by 36 gives 252:180 = 7:5. The ratio 5:7 is the reverse order, so it is not correct. Exam tip: Keep the order of the sequences the same while writing a ratio.

This is an arithmetic progression with first term 12 and common difference 9. Its first 5 terms are 12, 21, 30, 39, and 48, and their sum is 12+21+30+39+48=150. The sum of the first 4 terms is only 102, so 4 is not correct. Exam tip: when needed, use \(S_n=\frac{n}{2}[2a+(n-1)d]\) for the sum of the first n terms of an AP.

Each term in this sequence is obtained by adding 11 to the previous term: 14, 25, 36, 47, 58, 69, 80, 91. Therefore, the term immediately before 91 is 80. Although 82 may seem close, it is not a term of this sequence. Exam tip: identify the common difference first, then subtract it once from the given term.

Each term in the sequence is half of the preceding term: \(768, 384, 192, 96, 48, 24\). Therefore, the term immediately before \(24\) is \(48\). Values such as \(36\) and \(54\) do not follow the halving rule. Exam tip: check the ratio of consecutive terms; here it is \(\frac{1}{2}\).

This is an arithmetic progression with first term \(a=19\) and common difference \(d=27-19=8\). Therefore, \(a_n=a+(n-1)d=19+(n-1)\times8=8n+11\). In option B, substituting \(n=1\) gives 27, so it does not produce the first term. Exam tip: for the \(n\)th term of an AP, first identify \(a\) and \(d\), then use \(a+(n-1)d\).

This is an arithmetic progression with first term \(a_1=156\) and common difference \(d=144-156=-12\). Therefore, \(a_n=a_1+(n-1)d=156-12(n-1)=168-12n\). In \(a_n=156-12n\), putting \(n=1\) gives 144, so it cannot represent the first term. Exam tip: Test an nth-term rule by substituting \(n=1\).

Each new term equals the sum of the two preceding terms. Thus the sequence is 9, 16, 25, 41, 66, 107. Therefore, the sixth term is 107. The option 100 is incorrect because after the fifth term is 66, the next term must be 66 + 41 = 107. Exam tip: Write recursive sequence terms in order and count their positions carefully.

This is a geometric sequence in which each term is obtained by multiplying the previous term by 5. Therefore, its nth term is \(a_n=8\times5^{n-1}\). From \(8\times5^{n-1}=5000\), we get \(5^{n-1}=625=5^4\). Hence, \(n-1=4\) and \(n=5\). At position 4, the term is 1000, not 5000. Exam tip: First identify the common ratio, then use the nth-term formula to find the position.

(a_6=5(6^2)-4(6)+1=157) and (a_3=5(3^2)-4(3)+1=34). So the difference is (157-34=123), not listed in the options.

Using the stated rule, double the last term 61 and add 1: \(2\times61+1=123\). Therefore, the next term is 123. Note that the earlier listed terms do not fully follow this rule, but applying the rule explicitly given in the question gives 123. Exam tip: When a rule is stated, apply it to the last given term to find the next term.

(a_{10}=10^2+2\times10=120), so the correct answer is (120). Exam tip: substitute the given position directly for (n).

Given \(a_n=n(n+1)=72\). So \(n^2+n-72=0\), i.e. \((n+9)(n-8)=0\). Since a term number must be positive, \(n=8\). Check: \(a_8=8\times9=72\). The 9th term is \(9\times10=90\), so it is not correct. Exam tip: for a product of consecutive integers, look for nearby factor pairs of the given number.

The differences between consecutive terms are \(11-5=6\), \(19-11=8\), \(29-19=10\), and \(41-29=12\). These differences increase by 2 each time, so the next difference is \(14\). Therefore, the next term is \(41+14=55\). Choosing 54 would give a difference of 13, which does not follow the pattern. Exam tip: For a sequence, first list consecutive differences and check their pattern.