Sequences

The terms are consecutive cube numbers: \(1=1^3\), \(8=2^3\), \(27=3^3\), and \(64=4^3\). Hence, the \(n\)th term is \(n^3\), so the \(6\)th term is \(6^3=216\). \(343\) is a close distractor, but it equals \(7^3\). Exam tip: Write the first few terms as perfect cubes to identify the pattern quickly.

This is a triangular-number sequence: each new term is obtained by adding 1, 2, 3, 4, and so on. After 15, the next terms are 15+6=21 and 21+7=28. Therefore, the 7th term is 28. Note that 21 is the 6th term, so it is a close but incorrect option. Exam tip: inspect the differences between consecutive terms to identify the pattern.

From the third term onward, each term is the sum of the two immediately preceding terms: \(2+3=5\), \(3+5=8\), and \(5+8=13\). Therefore, the next term is \(8+13=21\). Option 20 is not correct because it is not the sum of the previous two terms. Exam tip: When finding a sequence rule, check sums of earlier terms as well as differences between consecutive terms.

From the third term onward, each term is the sum of the two immediately preceding terms: 11 = 4 + 7, 18 = 7 + 11, and 29 = 11 + 18. Therefore, the next term is 18 + 29 = 47. The number 45 would result from adding 16 to 29, but that does not follow the sum rule. Exam tip: In such sequences, check the sum of the previous two terms as well as consecutive differences.

Each term in this sequence is twice the previous term: 6, 12, 24, 48, 96, 192. Therefore, 192 is the sixth term. The fifth term is 96, so it is a close but incorrect option. Exam tip: List the terms in order or keep multiplying by 2 to identify the position of a term.

This is an arithmetic sequence with first term 9 and common difference 5. Therefore, its nth term is \(a_n=9+(n-1)\times5\). Substituting \(a_n=49\), we get \(9+(n-1)\times5=49\), so \(n-1=8\) and \(n=9\). Hence, 49 is the 9th term. The 10th term would be 54, so it is not correct. Exam tip: To find the position of a given term, substitute it in \(a_n=a+(n-1)d\) and solve for \(n\).

This is an arithmetic sequence with first term 75 and common difference \(-6\). Thus, \(a_n=75+(n-1)(-6)\). Putting \(a_n=39\), we get \(75-6(n-1)=39\), so \(n-1=6\) and \(n=7\). Therefore, 39 is the 7th term. The 6th term is 45, so it is a close but incorrect option. Exam tip: use \(a_n=a+(n-1)d\) to find the position of a term.

The differences between consecutive terms are \(7-2=5\), \(12-7=5\), and \(17-12=5\), so this is an arithmetic sequence with common difference \(d=5\). Since \(a_1=2\), \(a_n=a_1+(n-1)d=2+5(n-1)=5n-3\). In option B, putting \(n=1\) gives 7, so it is incorrect. Exam tip: check the common difference first, then verify a rule by substituting \(n=1\).

This is an arithmetic sequence because the difference between consecutive terms is 7. Thus, \(a_n=a+(n-1)d=4+(n-1)\times7=7n-3\). Therefore, option A is correct. In option B, putting \(n=1\) gives the first term as 11, not 4. Exam tip: Check a proposed rule by substituting \(n=1\) and then \(n=2\).

For the first four terms, substitute n=1, 2, 3, and 4. This gives a_1=2(1)+3=5, a_2=7, a_3=9, and a_4=11. Therefore, the sequence is (5, 7, 9, 11). Option A results from effectively omitting the constant 3. Exam tip: always check the starting value of n before listing terms.

To find the fourth term, substitute \(n=4\) in the rule: \(a_4=4^2+2(4)=16+8=24\). Therefore, \(24\) is correct. A choice such as \(22\) may result from evaluating \(4^2\) incorrectly. Exam tip: evaluate powers first, then multiplication and addition.

For the third term, substitute \(n=3\) in the rule: \(a_3=50-4(3)=50-12=38\). Hence, 38 is correct. Getting 36 would result from an incorrect multiplication or subtraction. Exam tip: for the \(k\)th term of a sequence, always replace \(n\) with \(k\).

The rule is \(a_n=3n^2\). Substituting \(n=1,2,3,4\) gives \(3\times1^2=3\), \(3\times2^2=12\), \(3\times3^2=27\), and \(3\times4^2=48\). Hence, the correct sequence is \(3,12,27,48\). Option C contains only the values of \(n^2\); they have not been multiplied by 3. Exam tip: substitute small values of \(n\) to check the first few terms of a sequence.

Starting with \(n=1\), \(a_1=2^1+1=3\). Similarly, \(a_2=5\), \(a_3=9\), and \(a_4=17\). Hence, the correct sequence is \(3, 5, 9, 17\). In option A, the first term is taken as 2, but the formula requires adding 1 after each power of 2. Exam tip: To find the first four terms, substitute \(n=1,2,3,4\) in order.

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.

A sequence has a common difference only when subtracting each term from the next produces the same result every time. For option A, the differences are 3, 3, and 3, so it is arithmetic. For option B, they are -4, -4, and -4. For option D, they are 6, 6, and 6. However, option C gives 9 - 4 = 5, 16 - 9 = 7, and 25 - 16 = 9. Since 5, 7, and 9 are not equal, option C does not have a common difference and is the correct answer. Its terms are successive squares, but that fact is not needed for the test.

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 \(a_n=a+(n-1)d\).