Sequences

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.

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.