Sequences

Each term in this sequence increases by 4, so the first term is 3 and the common difference is 4. Thus, \(a_{10}=3+(10-1)\times4=3+36=39\). The value 37 is obtained after adding 4 only eight times, so it is the 9th term. Exam tip: for the \(n\)th term, add the common difference \(n-1\) times to the first term.

This is an arithmetic progression with first term \(a=42\) and common difference \(d=-5\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_8=42+(8-1)(-5)=42-35=7\). Hence, 7 is correct. Getting 12 usually results from counting one fewer decrease than required. Exam tip: for the \(n\)th term, count \(n-1\) common differences.

Given \(a_n=3n+4\), substitute \(n=7\) to find the seventh term: \(a_7=3\times7+4=21+4=25\). Hence, 25 is correct. The value 24 would result from using \(3\times7+3\), which does not match the given rule. Exam tip: For an \(a_n\) question, first substitute the required term number for \(n\), then simplify.

To find the fifth term, substitute n=5 in the formula: a_5=2(5)^2-1=2×25-1=49. Therefore, 49 is the correct answer. 51 may result from an error in the final subtraction. Exam tip: evaluate the power first, then multiply and subtract.

Check the difference between consecutive terms: 9 - 5 = 4 and 13 - 9 = 4. Thus, 4 is added to each term. Therefore, the term after 13 is 13 + 4 = 17, and 17 + 4 = 21 confirms the pattern. Option 16 is not correct because it would give 20 as the next term. Exam tip: In missing-term questions, first check the difference between consecutive terms.

Check the differences between the known terms: from 68 to 62 and from 62 to 56, the decrease is 6 each time. Thus, the rule is to subtract 6 for every next term. Therefore, the term after 80 is 80 - 6 = 74, and 74 - 6 = 68 also confirms the pattern. Choosing 72 would make the next difference 4, so it is incorrect. Exam tip: In missing-term sequences, first check the differences between consecutive given terms.

The differences between consecutive terms are \(6-2=4\), \(12-6=6\), \(20-12=8\), and \(30-20=10\). These differences increase by 2 each time, so the next difference is \(12\). Therefore, the next term is \(30+12=42\). Choosing \(40\) would repeat a difference of 10 instead of continuing the pattern. Exam tip: For such sequences, first list the consecutive differences and look for their pattern.

Look at the differences between consecutive terms: 2, 4, 6, and 8 are subtracted from 100. The subtracted even number increases by 2 each time, so the next subtraction is 10. Thus, 80 - 10 = 70. Option 72 would result from subtracting 8 again, but the subtraction pattern is 2, 4, 6, 8, 10. Exam tip: For such sequences, write the consecutive differences first and identify their pattern.

In this sequence, each term is twice the preceding term: 3, 6, 12, 24, 48, 96, 192. Therefore, the 7th term is 192. The number 96 is the 6th term, so it is a close but incorrect option. Exam tip: Count term positions starting with the first term as 1.

In this sequence, each term is obtained by dividing the preceding term by 3: \(243\div3=81\), \(81\div3=27\), and \(27\div3=9\). Therefore, the next term is \(9\div3=3\). Although 6 is obtained by subtracting 3 from 9, it does not follow the division rule of the sequence. Exam tip: First check for a multiplication or division relationship between consecutive terms.

This is a sequence of perfect squares: \(1^2, 2^2, 3^2, 4^2, 5^2, \ldots\). Hence, its \(n\)th term is \(n^2\). Therefore, the 9th term is \(9^2=81\). Note that 64 is \(8^2\), so it is the 8th term. Exam tip: for a sequence of perfect squares, square the term number to find the corresponding term.

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.