Sequences

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.

The given rule is \(a_n=5n-2\). To find the ninth term, substitute \(n=9\): \(a_9=5\times9-2=45-2=43\). Therefore, the correct answer is 43. The value 41 may result from an error in multiplication or subtraction. Exam tip: While finding an \(n\)th term, first substitute the given value of \(n\) carefully.

The given rule is \(a_n=2n^2+3\). Substituting \(n=6\), we get \(a_6=2\times6^2+3=2\times36+3=75\). Therefore, the correct answer is 75. The value 72 results if one calculates \(2\times6^2\) but forgets to add 3. Exam tip: In such questions, square \(n\) first, then multiply and add.

Check the differences between consecutive terms: 15 − 7 = 8 and 23 − 15 = 8. Thus, this is an arithmetic sequence in which 8 is added each time. Therefore, the term after 23 is 23 + 8 = 31, and 31 + 8 = 39 confirms the pattern. Choosing 29 would not maintain the common difference. Exam tip: For a missing term in a sequence, first check the differences between consecutive terms.

This is a decreasing sequence in which each term is 8 less than the previous term: \(110-8=102\) and \(102-8=94\). Therefore, the blank is 102. If 101 were used, the decrease from 110 would be 9, so the common difference would not remain the same. Exam tip: In a missing-term sequence, check the differences using terms on both sides of the blank.

The given terms are consecutive square numbers: \(4=2^2\), \(9=3^2\), \(16=4^2\), \(25=5^2\), and \(36=6^2\). Therefore, the next term is \(7^2=49\). Although 47 is close, it is not a perfect square. Exam tip: in a sequence of squares, the base numbers increase by 1 each time.

The successive subtractions are 4, 8, 12, and 16. Since each subtraction increases by 4, the next subtraction is 20. Therefore, the next term is \(110-20=90\). Choosing 88 would require subtracting 22, which does not follow the pattern. Exam tip: write the differences between consecutive terms to identify the pattern quickly.

Each term is 4 times the preceding term, so this is a geometric sequence. The sixth term is \(2\times4^{5}=2\times1024=2048\). Note that 1024 is the fifth term, making it a close but incorrect option. Exam tip: When consecutive terms have a common multiplier, use \(a_n=a_1r^{n-1}\).

Each term is obtained by dividing the preceding term by 5: \(625\div5=125\), \(125\div5=25\), and \(25\div5=5\). Therefore, the next term is \(5\div5=1\). Zero would result from subtracting 5 from 5, but the pattern here is division. Exam tip: Compare consecutive terms using division to identify a multiplication or division pattern.

This is a sequence of cube numbers: \(1=1^3\), \(8=2^3\), \(27=3^3\), \(64=4^3\), and \(125=5^3\). Hence, the \(n\)th term is \(n^3\). Therefore, the 7th term is \(7^3=343\). Although 216 is also a cube, it is \(6^3\), so it is the 6th term. Exam tip: Express the given terms as powers of natural numbers to identify the pattern quickly.

The successive differences are 3, 4, 5, 6, and so on. Therefore, adding 7 to 21 gives the sixth term 28; adding 8 gives the seventh term 36; and adding 9 gives the eighth term 45. Hence, 45 is correct. Note that 36 is the seventh term, not the eighth. Exam tip: For such sequences, first identify the pattern in the consecutive differences.