Sequences

The first (6) terms are (4,9,14,19,24,29). Their sum is (99), so no listed option is correct.

Each successive term in the sequence decreases by 6. Therefore, the first 5 terms are 60, 54, 48, 42, and 36. Their sum is 60+54+48+42+36=240, so 240 is correct. Note that 230 is not the sum of the first four terms either; those terms add up to 204. Exam tip: Write out the required terms before adding them in a decreasing sequence.

This is an arithmetic progression with first term \(a=2\) and common difference \(d=5\). Its \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(2+(n-1)\times5=47\), which gives \(n=10\). Therefore, 47 is the 10th term. The 9th term is \(42\), so it is not correct. Exam tip: To find the position of a given term, substitute it for \(a_n\) in \(a_n=a+(n-1)d\).

This is an arithmetic sequence in which each successive term increases by 4. Thus, the terms are 6, 10, 14, 18, 22, 26, \ldots Hence, \(a_5=22\) and \(a_6=26\). Therefore, \(a_5+a_6=22+26=48\), so option D is correct. The value 46 may result from calculating the sixth term incorrectly. Exam tip: while counting terms, treat the first term as \(a_1\).

This is an arithmetic progression with common difference \(d=21-15=6\). The third term is \(a_3=27\), and the sixth term is \(a_6=45\). Hence, \(a_6-a_3=45-27=18\). Option 12 represents the difference across only two terms, not three. Exam tip: use \(a_n-a_m=(n-m)d\) to find differences between terms quickly.

This is an arithmetic progression with common difference \(4\). Here, \(a_2=14\) and \(a_5=10+4\times4=26\). Therefore, \(a_5-a_2=26-14=12\). The value \(14\) is \(a_2\), not the difference between the two terms. Exam tip: for the difference of terms, subtract the lower-indexed term from the higher-indexed term.

This is an arithmetic progression with first term 9 and common difference 5. To reach the seventh term, 5 is added six times: \(a_7=9+(7-1)\times5=39\). Therefore, 39 is correct. Getting 44 would mean adding 5 seven times, which gives the eighth term. Exam tip: use \(a_n=a_1+(n-1)d\) for an arithmetic progression.

This is an arithmetic sequence with first term \(a_1=72\) and common difference \(d=-8\). To reach the sixth term, 8 is subtracted 5 times: \(a_6=72+5(-8)=72-40=32\). Therefore, 32 is correct. The nearby distractor 40 results from subtracting 8 only 4 times. Exam tip: before the \(n\)th term, there are always \(n-1\) common differences.

The recursive rule is to multiply the previous term by 3 and then add 1. Therefore, the term after 94 is \(94\times 3+1=282+1=283\). Although 282 is three times 94, it misses the required addition of 1. Exam tip: for a recursive sequence, apply the stated rule directly to the last given term.

The rule is to divide the previous term by 2 and then subtract 1. Therefore, the term after 6.25 is \(6.25\div 2-1=3.125-1=2.125\). Option 3.125 results from dividing by 2 only; the subtraction of 1 is still required. Exam tip: In recursive sequences, apply the operations in the stated order—divide first, then subtract.

The differences between consecutive terms are 3, 6, 9, and 12. Since each difference increases by 3, the next difference is 15. Therefore, the next term is \(35+15=50\). Choosing 49 would give a difference of 14, which does not follow the pattern. Exam tip: For such sequences, first list consecutive differences and check their pattern.

The consecutive terms decrease by 10, 20, 30, and 40 respectively. Therefore, the next decrease is 50: \(50-50=0\). Hence, the next term is 0. Option 10 would result from subtracting 40 from 50, but 40 has already been used as the previous decrease. Exam tip: For such sequences, write the differences between consecutive terms and look for their pattern.

This sequence also uses successive multiplication. The first term, 2, is multiplied by 2 to produce 4. Next, 4 is multiplied by 3 to produce 12, and 12 is multiplied by 4 to produce 48. The multiplying factors therefore follow the increasing pattern 2, 3, 4. The next factor should be 5.

Multiplying the last term by 5 gives \\(48\times5=240\\). Thus the next term is 240, which is option C. The sequence can also be described by \\(2\times1!, 2\times2!, 2\times3!, 2\times4!\\), producing 2, 4, 12, and 48; the next expression is \\(2\times5!=240\\). A value such as 288 would require a different rule not supported by the displayed pattern.

The fourth term of the first sequence is 192, and that of the second sequence is 320. Therefore, the ratio is 192:320. Dividing both terms by 64 gives 192:320 = 3:5, so option A is correct. The ratio 5:3 is the reverse ratio. Exam tip: While writing a ratio, keep the order of the first and second quantities unchanged.

This is an arithmetic progression with first term 8 and common difference 5. Its first 5 terms are 8, 13, 18, 23, and 28, and their sum is 8+13+18+23+28=90. The sum of the first 4 terms is only 62, so option 4 is not correct. Exam tip: For small numbers of terms, write the terms and add them to check quickly.

Each successive term in the sequence is obtained by adding 7: 6, 13, 20, 27, 34, 41. Therefore, the term immediately before 41 is 34. Option 36 is incorrect because adding 7 to 34 gives 41. Exam tip: Find the common difference between consecutive terms to identify a missing previous or next term quickly.

Each term in this sequence is half of the preceding term: \(160, 80, 40, 20, 10, 5\). Therefore, \(10\) comes immediately before \(5\). Values such as \(8\) and \(12\) do not follow the halving rule. Exam tip: Check the ratio of consecutive terms to identify a sequence rule quickly.

This is an arithmetic sequence with first term \(a=11\) and common difference \(d=15-11=4\). Therefore, \(a_n=a+(n-1)d=11+(n-1)\times4=4n+7\). For \(4n+11\), substituting \(n=1\) gives 15, so it does not produce the first term 11. Exam tip: use \(a_n=a+(n-1)d\) after identifying the first term and common difference.

This is an arithmetic sequence with first term \(a_1=60\) and common difference \(d=52-60=-8\). Therefore, \(a_n=a_1+(n-1)d=60+(n-1)(-8)=68-8n\). In option A, substituting \(n=1\) gives 60 and \(n=2\) gives 52. Option B gives 52 when \(n=1\), so it is incorrect. Exam tip: verify a proposed nth-term rule by checking \(n=1\) and \(n=2\).

Each new term equals the sum of the two preceding terms. Thus, \(a_3=4+9=13\), \(a_4=9+13=22\), and \(a_5=13+22=35\). Therefore, 35 is correct. The value 31 can result from adding terms in the wrong order. Exam tip: Write every term sequentially in recursive-sequence questions.

This is a geometric sequence in which each term is 4 times the preceding term. Hence, \(a_n=3\times4^{n-1}\). From \(3\times4^{n-1}=768\), we get \(4^{n-1}=256=4^4\). Therefore, \(n-1=4\) and \(n=5\). At position 4, the term is 192, not 768. Exam tip: first identify the common ratio when finding a term's position.

The rule is to multiply the previous term by 2 and then add 2. Therefore, the term after 30 is \(2\times 30+2=62\). Although 60 is double of 30, it does not include the required addition of 2. Exam tip: In a recursive sequence, apply every part of the rule to the last given term.

The differences between consecutive terms are \(12-5=7\), \(21-12=9\), and \(32-21=11\). These are consecutive odd numbers, so the next difference is \(13\). Therefore, the next term is \(32+13=45\). Choosing \(47\) would give a difference of \(15\), which does not continue the pattern. Exam tip: For such sequences, first list the consecutive differences and check their pattern.

This is an arithmetic sequence with first term \(a=6\) and common difference \(d=5\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{12}=6+(12-1)\times5=6+55=61\). Hence, 61 is correct. Choosing 56 would count only 10 differences, whereas there are 11 differences from the first term to the 12th term. Exam tip: always use \((n-1)\), not \(n\), in the nth-term formula.

This is an arithmetic progression because each successive term decreases by 7. Here, \(a=90\), \(d=-7\), and \(n=9\). Thus, \(a_9=a+(9-1)d=90+8(-7)=34\). Therefore, 34 is correct. The value 36 would result from subtracting 7 only seven times, but there are eight gaps from the first term to the ninth term. Exam tip: for the \(n\)th term, always use \(n-1\) common differences.