Sequences

From the third term onward, each term is the sum of the two immediately preceding terms: \(5+6=11\), \(6+11=17\), and \(11+17=28\). Therefore, the next term is \(17+28=45\). An option such as 43 does not follow the rule of adding the previous two terms. Exam tip: For such sequences, first check relations such as addition, subtraction, or multiplication between consecutive terms.

From the third term onward, each term is the sum of the two immediately preceding terms: 21 = 8 + 13, 34 = 13 + 21, and 55 = 21 + 34. Therefore, the next term is 34 + 55 = 89. A number such as 84 may result from extending the differences incorrectly. Exam tip: Verify the same rule using at least two consecutive terms before choosing the answer.

Each term in the sequence is 3 times the preceding term: 7, 21, 63, 189, 567. Therefore, 567 is the 5th term. Since 189 is the 4th term, the 4th-term option is incorrect. Exam tip: List the terms in order and count their positions carefully.

This is an arithmetic sequence with first term 12 and common difference 7. Its nth term is \(a_n=12+(n-1)\times7\). Putting \(a_n=75\), we get \(12+(n-1)\times7=75\), so \(n-1=9\) and \(n=10\). Therefore, 75 is the 10th term. The 9th term is \(68\), so option B is not correct. Exam tip: To find a term's position, identify the first term and common difference, then use \(a_n=a+(n-1)d\).

The statement is correct because consecutive differences are \(5-2=3\), \(8-5=3\), and \(11-8=3\). Hence, the term after 20 is \(20+3=23\). Exam tip: check the common difference first.

The difference between consecutive terms is 7, so this is an arithmetic sequence. Its nth term is \(a_n=a_1+(n-1)d=5+(n-1)\times7=7n-2\). For \(a_n=7n+5\), substituting \(n=1\) gives 12, not the first term 5. Exam tip: check a proposed rule by substituting \(n=1\) for the first term.

This is an arithmetic sequence with first term \(a_1=8\) and common difference \(d=18-8=10\). Hence, \(a_n=a_1+(n-1)d=8+(n-1)10=10n-2\). Therefore, option A is correct. In option B, substituting \(n=1\) gives the first term as \(18\), not \(8\). Exam tip: Test a proposed nth-term rule by substituting \(n=1\) and \(n=2\) to check the first two terms.

For the first four terms, substitute n=1, 2, 3, and 4. This gives a_1=6(1)+1=7, a_2=13, a_3=19, and a_4=25. Hence, the sequence is (7, 13, 19, 25). Option C incorrectly treats 1 as the first term; when n starts at 1, the first term is 7. Exam tip: List n=1, 2, 3, 4 and substitute each value in the nth-term rule.

To find the sixth term, substitute n=6 in the rule: a_6=6^2+4(6)=36+24=60. Therefore, 60 is correct. A value such as 54 usually results from evaluating 6^2 incorrectly or making an addition error. In exams, calculate powers first, then multiplication, and finally addition.

To find the fifth term, substitute \(n=5\) in the rule: \(a_5=90-7(5)=90-35=55\). Therefore, 55 is correct. Choosing 52 results from an error in multiplication or subtraction. Exam tip: In an \(n\)th-term rule, substitute the value of \(n\) first, using brackets, and then simplify.

Substituting n = 1, 2, 3, and 4 in the rule gives 5(1)^2 = 5, 5(2)^2 = 20, 5(3)^2 = 45, and 5(4)^2 = 80. Hence, the sequence is (5, 20, 45, 80). Option A consists of multiples of 5, but it follows the rule 5n, not 5n². Exam tip: Substitute the first few natural-number values of n to match a sequence with its rule.

Substitute n=1,2,3: a_1=4^1-2=2, a_2=4^2-2=14, and a_3=4^3-2=62. Hence, the correct sequence is (2,14,62). Option B lists only powers of 4 and does not subtract 2 from each term. Exam tip: Unless stated otherwise, begin finding sequence terms with n=1.

Both sequences have common difference (5), so the term difference always remains (7). The (8)th terms also differ by (7).

Both sequences have difference (7), and their first terms differ by (9). So their (7)th terms also differ by (9).

This is an arithmetic progression with first term 5 and common difference 7. Its first 6 terms are 5, 12, 19, 26, 33, and 40, whose sum is 135. A result such as 138 can arise from an error while adding the terms. Exam tip: You can also use \(S_n=\frac{n}{2}[2a+(n-1)d]\) directly for an arithmetic progression.

Each term in the sequence decreases by 7. Therefore, the first 5 terms are 72, 65, 58, 51, and 44. Their sum is 72+65+58+51+44=290, so option C is correct. A value such as 285 can result from an addition error. Exam tip: Write all the required terms in order before adding them.

This is an arithmetic progression with first term \(a=4\) and common difference \(d=6\). Its \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(4+(n-1)\times6=58\) gives \(n-1=9\), so \(n=10\). Therefore, 58 is the 10th term. The 9th term is \(52\), so it is not correct. Exam tip: To find the position of a given term in an AP, substitute that term for \(a_n\) in \(a_n=a+(n-1)d\).

This is an arithmetic sequence with first term 7 and common difference 5. Therefore, the sixth term is 32 and the seventh term is 37. Hence, \(a_6+a_7=32+37=69\). An option such as 64 results from using an incorrect term position or common difference. In exams, write the required terms separately before adding them.

This is an arithmetic sequence with common difference 7. Thus, \(a_4=39\) and \(a_7=18+6\times 7=60\). Therefore, \(a_7-a_4=60-39=21\). Option 18 may result from counting the gap between the terms incorrectly. Exam tip: In an arithmetic sequence, use \(a_n-a_m=(n-m)d\) to find the difference directly.

This is an arithmetic progression with common difference \(5\). Here, \(a_2=17\) and \(a_6=12+(6-1)\times5=37\). Therefore, \(a_6-a_2=37-17=20\). Choosing \(25\) incorrectly counts five intervals instead of the four intervals between the 2nd and 6th terms. Exam tip: use \(a_n-a_m=(n-m)d\) to find the difference between two terms.