Sequences

For the first three terms, substitute \(n=1,2,3\). This gives \(a_1=1^2-1=0\), \(a_2=2^2-1=3\), and \(a_3=3^2-1=8\). Hence, the correct sequence is \((0,3,8)\). Option \((1,4,9)\) uses only \(n^2\) and does not subtract 1. Exam tip: to find initial terms from an nth-term rule, substitute 1, 2, and 3 in order and check each calculation.

For the first term, take \(n=1\). Thus, \(a_1=2^1=2\), \(a_2=2^2=4\), \(a_3=2^3=8\), and \(a_4=2^4=16\). Hence, the correct sequence is \((2, 4, 8, 16)\). Option A starts with \(2^0\), but the first term conventionally corresponds to \(n=1\). Exam tip: To list the first four terms, substitute \(n=1,2,3,4\) in the given rule.

This is an arithmetic sequence because 4 is added to each successive term. The terms are 8, 12, 16, 20, 24, 28; therefore, the 6th term is 28. Although 24 is a close distractor, it is the 5th term. Exam tip: write the terms in order and count their positions carefully.

In this sequence, each successive term is 6 less than the previous one: 90, 84, 78, 72, 66, 60. Therefore, the 6th term is 60. The number 66 is the 5th term, so it is a close but incorrect option. Exam tip: Count the first listed term as the 1st term.

Each term of this sequence is a multiple of 5, so its nth term is \(5n\). Putting \(5n=35\) gives \(n=7\); hence, 35 is the 7th term. The 6th term is \(30\), so the closest distractor is incorrect. Exam tip: To find a term’s position, equate the general term to the given value.

This is an arithmetic sequence with first term 3 and common difference 4. Its terms are 3, 7, 11, 15, 19; therefore, 19 is the fifth term. The fourth term is 15, so the fourth option is not correct. Exam tip: list the terms in order and count their positions.

Each successive term is 4 less than the preceding term: 48, 44, 40, 36, 32. Hence, 32 is the fifth term. The fourth term is 36, so the fourth-term option is incorrect. Exam tip: To find a term’s position, write the sequence with term numbers 1, 2, 3, and so on.

In the sequence (6, 12, 18, 24), the first term is 6, the second is 12, and the third is 18. Therefore, option A is correct. In option C, 18 is the second term, not the third. In exams, count terms from the left to identify their positions.

In the sequence (7, 12, 17, 22), the first term is 7 and the consecutive differences are 12 - 7 = 5, 17 - 12 = 5, and 22 - 17 = 5. Therefore, option B is correct. In option A, 4 is added each time, not 5. In exams, check both the first term and the difference between consecutive terms.

In the sequence \(64, 56, 48, 40\), the differences between consecutive terms are \(56-64=-8\), \(48-56=-8\), and \(40-48=-8\). Thus, 8 is subtracted each time. In option D, the terms decrease by 16, so it is not correct. Exam tip: Find the difference between consecutive terms to check a sequence rule quickly.

The terms of the sequence are 13, 17, 21, 25, \(\ldots\). Here, \(a_3\) denotes the third term, so \(a_3=21\). The number 25 is the fourth term, not the third. Exam tip: the subscript \(n\) in \(a_n\) indicates the position of the term.

The terms of the sequence are 70, 63, 56, and 49 in order. Therefore, the fourth term, \(a_4\), is \(49\). The number \(56\) is the third term, so it cannot be \(a_4\). Exam tip: To identify \(a_n\), count the terms from the left as 1, 2, 3, and so on.

In a sequence, \(a_n\) denotes the \(n\)th term. Hence, \(a_2\) is the second term. The given terms are 2, 11, 20, and 29, so \(a_2=11\). The number 20 is the third term, \(a_3\), not the second term. Exam tip: Count terms from the first term when reading sequence notation.

In a sequence, \(a_1\) denotes the first term. The first term of \(18,24,30,36,\ldots\) is 18, so 18 is correct. Here, 24 is the second term, \(a_2\), not \(a_1\). Exam tip: \(a_1\), \(a_2\), and \(a_3\) represent the first, second, and third terms respectively.

The differences between consecutive terms are 2, 3, 4, and 5. Since each difference increases by 1, the next difference is 6. Therefore, the next term is \(15+6=21\). Choosing 20 would give a difference of 5, but the required next difference is 6. Exam tip: For such sequences, list consecutive differences first and look for their pattern.

The differences between consecutive terms are 3, 5, 7, and 9, which are consecutive odd numbers. Therefore, the next difference is 11. Hence, the next term is \(26+11=37\). Choosing 35 would give a difference of only 9 and would break the pattern of increasing odd differences. Exam tip: For such sequences, first write the consecutive differences to identify the pattern.

The differences between consecutive terms are \(7-5=2\), \(10-7=3\), \(14-10=4\), and \(19-14=5\). Since the difference increases by 1 each time, the next difference is \(6\). Therefore, the next term is \(19+6=25\). Option 24 would result if the next difference remained 5, which does not follow the pattern. Exam tip: Write the consecutive differences first to identify a sequence pattern.

In this sequence, each term is obtained by dividing the previous term by 2: 120 to 60, 60 to 30, and 30 to 15. Therefore, the next term is \(15 \div 2 = 7.5\). Option 8 is not half of 15. Exam tip: Check for a common multiplication or division rule between consecutive terms.

To find the third term, substitute \(n=3\) in the rule: \(a_3=6\times3+1=18+1=19\). Hence, 19 is the correct answer. The value 18 is only \(6\times3\), so it misses the \(+1\) in the rule. Exam tip: substitute the given value of \(n\) first, then carry out all operations carefully.

Given \(a_n=3n^2\), substitute \(n=2\): \(a_2=3\times2^2=3\times4=12\). Hence, option A is correct. Option B results from multiplying \(3\times2\) without squaring 2. Exam tip: while finding a term of a sequence, substitute the given value of \(n\) first and then evaluate the exponent.

To find the first three terms, substitute n=1, 2, and 3. This gives a_1=15-2(1)=13, a_2=15-2(2)=11, and a_3=15-2(3)=9. Therefore, the sequence begins with (13, 11, 9). The option (15, 13, 11) would result if indexing started at n=0. Exam tip: Unless stated otherwise, use n=1 to find the first term from a_n.

Each term in the sequence is twice the preceding term: 10, 20, 40, 80. Therefore, the next term is \(80\times 2=160\). The value 120 would result from adding 40 to 80, but the pattern is multiplication by 2, not addition. Exam tip: check the ratio of consecutive terms; here \(20/10=40/20=80/40=2\).

Write the sequence in pairs: (2, 4), (3, 6), (4, 8). In each pair, the second term is twice the first term, while the first terms 2, 3, 4, ... increase by 1. Therefore, the term after 8 is 5. Option 10 is not the next term; it would come after 5. Exam tip: In such sequences, examine the odd-position and even-position terms separately to identify the pattern.

The differences between consecutive terms are \(2, 4, 2, 4\). The pattern alternates between \(+2\) and \(+4\), so add \(+2\) after 13: \(13+2=15\). Therefore, 15 is the correct answer. Option 17 would result from adding 4 next, but the next difference in the pattern is 2. Exam tip: For next-term questions, first write the differences between consecutive terms to identify the pattern.

The governing concept is an arithmetic sequence, whose consecutive terms have one constant difference. Check each option by subtracting every term from the next. In A, the differences are 10−7=3, 13−10=3, and 16−13=3, so the difference is constant. In B, they are 16−20=−4, 12−16=−4, and 8−12=−4. In D, they are 11−5=6, 17−11=6, and 23−17=6. In C, however, the differences are 9−4=5, 16−9=7, and 25−16=9. Since these are unequal, C is not an arithmetic sequence and is correct. Its terms are consecutive squares, but that observation is not required.