Sequences

In the sequence (5, 9, 13, 17), the differences between consecutive terms are 9−5=4, 13−9=4, and 17−13=4. Hence, its common difference is 4, so it is an arithmetic sequence. In (1, 3, 6, 10), the differences are 2, 3, and 4, which are not equal. Exam tip: subtract consecutive terms to check for a common difference.

In the sequence (18, 15, 12, 9), the differences between consecutive terms are 15−18 = −3, 12−15 = −3, and 9−12 = −3. Hence, 3 is subtracted each time. In option B, 4 is subtracted each time, so it is not correct. Exam tip: Subtract consecutive terms to check the common difference.

In option D, the terms are \(1^2, 2^2, 3^2, 4^2\), respectively, so it is a sequence of square numbers. Option A is a sequence of natural numbers, while B and C are multiples of 2 and 3 respectively. Exam tip: check whether each term can be written as \(n^2\).

Odd numbers are not exactly divisible by 2, such as 1, 3, 5, and 7. Therefore, (1, 3, 5, 7) is a sequence containing only odd numbers. In option A, every term is even because it is divisible by 2. Exam tip: A whole number ending in 1, 3, 5, 7, or 9 is odd.

Even numbers are divisible by 2 without a remainder. Every term in (2, 4, 6, 8) is divisible by 2, so it is a sequence of even numbers. In options B and D, some terms are even, but 3 and 5 are odd. Exam tip: Check every term of a sequence; having only some even terms is not enough.

The rule is \(a_n=n^2\). For the fourth term, substitute \(n=4\): \(a_4=4^2=16\). Option 8 is a common error caused by multiplying 4 by 2 instead of squaring it. Exam tip: In any \(a_n\), first substitute the term number for \(n\), then simplify.

The given rule is (a_n=5n). For the sixth term, put n=6: (a_6=5×6=30). Therefore, the correct answer is 30. The value 25 is the fifth multiple of 5, so it represents (a_5), not (a_6). Exam tip: To find the nth term, substitute the required value of n directly into the given rule.

The terms 2, 4, 6, 8 are written in a definite order, so they form a sequence. Having only four terms makes it a finite sequence; infinity is not required. Exam tip: first check whether the terms have an order.

Given \(a_n=2n+1\), substitute \(n=5\) to find the fifth term: \(a_5=2(5)+1=10+1=11\). Therefore, 11 is correct. The close distractor 10 results from forgetting to add \(+1\). Exam tip: substitute the term number into the rule before simplifying.

This is an arithmetic sequence with first term 4 and common difference 3. The seventh term is \(a_7=4+(7-1)\times3=22\). The number 21 is the sixth term, so it is a close but incorrect option. Exam tip: to find the \(n\)th term, add the common difference \(n-1\) times.

Each successive term is obtained by subtracting 5: 100, 95, 90, 85, 80, 75. Therefore, the sixth term is 75. The closest distractor, 80, is the fifth term, not the sixth. Exam tip: Write the terms in order and count carefully in sequence questions.

Each successive term in the sequence increases by 4: 2, 6, 10, 14, 18. Therefore, 18 is the fifth term. The fourth term is 14, so the fourth position is not correct. Exam tip: Count the terms in order, or use the AP formula \(a_n=a+(n-1)d\) to find a term's position.

Each successive term decreases by 5: 50, 45, 40, 35, 30. Hence, 30 is the fifth term. The fourth term is 35, so the fourth option is not correct. In exams, count the first given number as the first term.

The terms of this sequence are consecutive multiples of 6: \(6\times1, 6\times2, 6\times3,\ldots\). Since \(36=6\times6\), 36 is the sixth term. The fifth term is \(6\times5=30\), so it is not correct. Exam tip: In such a sequence, divide the given term by 6 to find its position.

In the sequence (4, 8, 12, 16), the terms are 4, 8, 12, and 16 in order, so 12 is the third term. In option C, 12 is the second term, while in options B and D it is the first and fourth term respectively. In exams, count terms from left to right to identify their positions.

In the sequence (5, 9, 13, 17), adding 4 to 5 gives 9, then adding 4 gives 13 and 17. Thus, its first term is 5 and its common difference is 4. Option A starts with 5, but it increases by 3 each time. Exam tip: always check both the first term and the difference between consecutive terms.

The correct sequence is (30, 24, 18, 12), because subtracting 6 from 30 gives 24, then subtracting 6 gives 18, and then 12. Its common difference is \(-6\). In option B, 5 is subtracted each time, so it is not correct. Exam tip: Check the difference between consecutive terms to identify the rule.

For the first four terms, substitute n=1, 2, 3, and 4 respectively. This gives a_1=1+2=3, a_2=2+2=4, a_3=3+2=5, and a_4=4+2=6. Hence, the sequence is (3,4,5,6). Option B starts with 2, which would result from using n=0; however, a sequence usually starts with n=1 unless stated otherwise. Exam tip: To find initial terms from an nth-term rule, substitute consecutive values n=1, 2, 3, ....

For the first four terms, substitute n=1, 2, 3, and 4. This gives a₁=2(1)-1=1, a₂=3, a₃=5, and a₄=7. Hence, the sequence is (1, 3, 5, 7). Option (2, 4, 6, 8) is the sequence of even numbers, so it does not follow the given rule. Exam tip: When finding initial terms from an nth-term rule, usually begin with n=1.

For the first four terms, take n=1,2,3,4. Using a_n=3n gives 3×1=3, 3×2=6, 3×3=9, and 3×4=12. Therefore, the correct sequence is (3,6,9,12). Option D also has a common difference of 3, but its first term is 4, so it does not follow this rule. Exam tip: For the first term of an nth-term rule, always substitute n=1.

Substitute \(n=1, 2, 3, 4\) in the rule: \(a_1=1^2=1\), \(a_2=2^2=4\), \(a_3=3^2=9\), and \(a_4=4^2=16\). Hence, the correct sequence is \((1, 4, 9, 16)\). Option D is the sequence of triangular numbers, not square numbers. Exam tip: to find initial terms, substitute successive natural-number values of n into the formula.

To find the first term, substitute \(n=1\) in the rule: \(a_1=20-2(1)=18\). Therefore, the correct answer is 18. Choosing 20 would be incorrect because it is the constant term before subtracting \(2n\). Exam tip: To find a particular term from \(a_n\), substitute that term’s index for \(n\).

In a sequence, \(a_n\) denotes the nth term. Therefore, \(a_2\) is the second term. The terms given are 12, 15, 18, and 21, so \(a_2=15\). Note that 18 is the third term, \(a_3\). Exam tip: use the subscript to identify the position of the term in the sequence.

The terms of the sequence are 9, 18, 27, and 36 in order. The notation \(a_4\) denotes the fourth term, so \(a_4=36\). Since 27 is the third term, it is not correct. Exam tip: the subscript in \(a_n\) gives the position of the term.

In a sequence, \(a_1\) denotes the first term. The first term of \(5,15,25,35,\ldots\) is 5, so \(a_1=5\). The number 15 is the second term, \(a_2\), so it is not correct. Exam tip: the subscript in \(a_n\) indicates the position of the term.