Sequences

Each term of the first sequence increases by 3, so its 6th term is \(2+5\times3=17\). The 6th term of the second sequence is \(1+5\times3=16\). Therefore, the difference is \(17-16=1\). The nearby option 0 is incorrect because corresponding terms of the two sequences differ by 1. Exam tip: for two sequences with the same common difference, the difference between corresponding terms remains constant.

Both sequences have difference (5), so the term difference always remains (6). The (8)th terms also differ by (6).

This is an arithmetic sequence with first term 3 and common difference 5. Its first 5 terms are 3, 8, 13, 18, and 23. Therefore, \(3+8+13+18+23=65\). A result such as 60 can arise from adding a term incorrectly. Exam tip: write the required terms first, then check the addition.

The first four terms are 40, 36, 32, and 28 because each successive term decreases by 4. Therefore, \(40+36+32+28=136\). A result of 140 can arise from calculating the fourth term incorrectly. Exam tip: Write all the required terms before adding them.

This is an arithmetic progression with first term 1 and common difference 3. Its nth term is \(a_n=1+(n-1)\times3\). Putting \(a_n=22\), we get \(1+(n-1)\times3=22\), so \(n=8\). Hence, 22 is the 8th term. The 9th term is 25, so it cannot be correct. Exam tip: To find the position of a given term, substitute its value in \(a_n=a+(n-1)d\).

This is an arithmetic progression with first term 5 and common difference 3. Hence, the sixth term is 20 and the seventh term is 23. Therefore, \(a_6+a_7=20+23=43\). Choosing 45 would result from an error in finding or adding the terms. Exam tip: Write the terms in order first and check their positions carefully.

Each successive term in the sequence increases by 6. Thus, the second term is \(a_2=18\) and the fifth term is \(a_5=36\). Therefore, \(a_5-a_2=36-18=18\). Option 24 is the fourth term, not the difference between the required terms. Exam tip: identify the required term positions first, then subtract their values.

In the sequence, the first term is \(a_1=9\) and the fourth term is \(a_4=21\). Therefore, \(a_4-a_1=21-9=12\), so \(12\) is correct. Choosing \(14\) would give an incorrect difference between the terms. Exam tip: subtract the lower-indexed term from the higher-indexed term when finding \(a_n-a_m\).

This is an arithmetic progression with first term 6 and common difference 4. To reach the sixth term, 4 is added 5 times: \(a_6=6+(6-1)\times4=26\). Therefore, 26 is correct. Getting 24 would mean adding the difference only 4 times. Exam tip: use \(a_n=a_1+(n-1)d\) to find the \(n\)th term.

Option C ends at \(20\), so it has exactly four terms and is a finite sequence. In options A, B, and D, \(\ldots\) indicates that the terms continue, so they are infinite sequences. Exam tip: a sequence with a definite last term is finite.

The rule is: each term equals twice the previous term plus 1. Hence, the term after 23 is \(2\times 23+1=46+1=47\). Option 46 is only double of 23; it misses the additional 1. In exams, 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 11 is \(11\div 2-1=5.5-1=4.5\). The value 5.5 is obtained only after dividing 11 by 2; subtracting 1 is still required. Exam tip: In recursive sequences, follow the stated order of operations carefully.

The differences between consecutive terms are 6−4=2, 10−6=4, 16−10=6, and 24−16=8. These differences increase by 2 each time, so the next difference is 10. Therefore, the next term is 24+10=34. Choosing 32 would repeat a difference of 8, whereas the difference pattern should be 2, 4, 6, 8, 10. Exam tip: For such sequences, first list consecutive differences and look for their pattern.

The differences between consecutive terms are 3, 6, 9, and 12. Since the amount subtracted increases by 3 each time, the next subtraction is 15. Therefore, \(60-15=45\). Option 42 would require subtracting 18, which does not follow the pattern. Exam tip: Write the differences between consecutive terms first to identify the pattern.

The sequence follows a multiplying pattern rather than a constant-addition pattern. Starting with 1, multiply by 2 to get 2. Then multiply by 3 to get 6, and multiply by 4 to get 24. The multipliers increase successively as 2, 3, and 4, so the next multiplier is 5.

Multiplying the last known term by this next factor gives \\(24\times5=120\\). Therefore the next term is 120, which is option C. Another way to view the sequence is through factorials: \\(1=1!\\), \\(2=2!\\), \\(6=3!\\), and \\(24=4!\\), so the next term is \\(5!=120\\). Choices such as 96 do not continue the observed multiplication pattern.

The fourth term of the first sequence is \(54\), and that of the second sequence is \(135\). Therefore, the ratio is \(54:135\). Dividing both terms by \(27\) gives \(54:135=2:5\). \(5:2\) is the reverse of the required ratio. Exam tip: Keep the order of terms in a ratio the same as the order of the sequences in the question.

This is an arithmetic progression with first term 6 and common difference 4. Its first 5 terms are 6, 10, 14, 18, and 22, and their sum is 6 + 10 + 14 + 18 + 22 = 70. Therefore, the correct answer is 5. The sum of the first 4 terms is only 48. Exam tip: use \(S_n=\frac{n}{2}[2a+(n-1)d]\) for the sum of the first n terms of an arithmetic progression.

Each term in this sequence is obtained by adding 5 to the previous term: 4, 9, 14, 19, 24, 29, 34. Therefore, the term immediately before 34 is 29. Option 31 is not a term of this sequence. Exam tip: Identify the common difference and move backward from the given term when asked for the preceding term.

Each term in the sequence is half of the preceding term: 96, 48, 24, 12, 6, 3. Therefore, the term immediately before 3 is 6. Although 12 occurs earlier in the sequence, 6 comes between 12 and 3, so 12 is not the immediate preceding term. Exam tip: identify the pattern and extend the sequence to find a required next or previous term.

This is an arithmetic progression with first term \(a=7\) and common difference \(d=3\). Therefore, \(a_n=a+(n-1)d=7+(n-1)\times3=3n+4\). The close distractor \(3n+7\) gives \(10\) when \(n=1\), so it cannot represent this sequence. 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 \(a_1=50\) and common difference \(d=-5\). Therefore, \(a_n=a_1+(n-1)d=50+(n-1)(-5)=55-5n\). In option A, \(n=1\) gives 50 and \(n=2\) gives 45. Option B gives 45 when \(n=1\), so it is incorrect. Exam tip: In \(a_n=a_1+(n-1)d\), do not forget the \(n-1\) term.

The rule says that each new term equals the sum of the two immediately preceding terms. Thus, \(a_3=3+8=11\), \(a_4=8+11=19\), and \(a_5=11+19=30\). Therefore, the correct answer is 30. Note that 19 is the fourth term, not the fifth. Exam tip: In a recursive sequence, write the terms in order and check their subscripts carefully.

Each term is obtained by multiplying the previous term by 3: 2, 6, 18, 54, 162. Therefore, 162 is the fifth term, so \(n=5\). Since 54 is the fourth term, 4 is a close but incorrect option. Exam tip: identify the multiplication pattern before counting the terms.

This is an arithmetic sequence because 6 is added to each successive term. Here, the first term is \(a=4\) and the common difference is \(d=6\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_9=4+(9-1)\times6=4+48=52\). Hence, 52 is the correct answer. A common error leading to 50 is counting the gaps between terms incorrectly. Exam tip: from the first term to the \(n\)th term, there are \(n-1\) common differences.

This is an arithmetic sequence because each successive term decreases by 6. Here, the first term is \(a=64\) and the common difference is \(d=-6\). Therefore, \(a_7=a+(7-1)d=64+6(-6)=28\). Choosing 30 usually results from counting the number of subtractions incorrectly. Exam tip: for the \(n\)th term, use \(a_n=a+(n-1)d\).