Explicit or general rule

The general rule gives each term by substituting its position for \(n\). Putting \(n=2\), we get \(a_2=3^2=9\). Option 8 is incorrect because it equals \(2^3\), whereas the base here is 3 and the exponent is 2. Exam tip: first substitute the term number for \(n\), then evaluate the power.

This is a geometric sequence in which each term is 3 times the previous term. Starting with \(n=1\), we get \(a_1=3^1=3\), \(a_2=3^2=9\), \(a_3=3^3=27\), and \(a_4=3^4=81\). Hence, the correct general term is \(a_n=3^n\). The rule \(a_n=3n\) would give 3, 6, 9, 12, so it is not correct. Exam tip: Substitute \(n=1\) in a proposed rule to check the first term.

Given \(a_n=n^2-n\), substitute \(n=5\): \(a_5=5^2-5=25-5=20\). Hence, \(20\) is correct. \(25\) is only the value of \(5^2\); subtracting \(5\) is also required. Exam tip: substitute the value of \(n\) in every occurrence in the formula.

For term numbers \(n=1,2,3,4\), the rule \(n^2-n\) gives \(0,2,6,12\), respectively. Therefore, the correct general term is \(a_n=n^2-n\). The rule \(a_n=2n\) gives 2 as its first term, so it does not match the sequence. Exam tip: verify a general term by substituting \(n=1\) and \(n=2\) and checking the first two terms.

Since \(a_n=7\), every term remains 7 regardless of \(n\), so it is a constant sequence. Its common difference is 0, not 7. Exam tip: list terms as 7, 7, 7 to identify it quickly.

The sequence follows an explicit arithmetic rule. Its common difference is 11 − 7 = 4, so the coefficient of n in the general term is 4. Applying a_n = a_1 + (n − 1)d gives a_n = 7 + (n − 1)4 = 7 + 4n − 4 = 4n + 3. Therefore option A is correct. Substitution confirms that n = 1 gives 7, n = 2 gives 11, n = 3 gives 15, and n = 4 gives 19. Option B gives 7, 14, 21, …; option C starts with 1; and option D gives 7, 8, 9, …. These alternatives either use the wrong difference or fail to preserve the pattern, so they cannot be the required general rule.

In \(a_n=5n+2\), the coefficient of \(n\) is 5, so each successive term increases by 5. Hence it is an AP, not a GP. Exam tip: in the form \(pn+q\), the common difference is \(p\).

The difference is (8), and at (n=1) the value must be (5), so (a_n=8n-3). In exams, find the constant using the first term.

Given \(a_n=\frac{n}{2}\). For the sixth term, substitute \(n=6\): \(a_6=\frac{6}{2}=3\). Therefore, the correct answer is 3. Choosing \(6\) would ignore the division by 2 in the formula. Exam tip: to find a particular term, substitute its subscript into the general rule.

This is an arithmetic progression with first term \(\frac{1}{2}\) and common difference \(\frac{1}{2}\). Therefore, \(a_n=\frac{1}{2}+(n-1)\frac{1}{2}=\frac{n}{2}\). Check: for \(n=1\), \(a_1=\frac{1}{2}\), and for \(n=2\), \(a_2=1\). The rule \(a_n=n\) gives the first term as 1, so it is incorrect. Exam tip: substitute the first two or three values of \(n\) to verify a general term.

Substituting \(n=1,2,3,4\) in \(2n^2\) gives \(2,8,18,32\), respectively. Hence, the correct general term is \(a_n=2n^2\). The close distractor \(a_n=4n\) gives 8 for the second term, but it gives 4 rather than 2 when \(n=1\). Exam tip: verify a proposed general term using at least the first three terms.

From (3n-2=13), we get (n=5). In exams, when term number is asked, equate the formula to the given value.

Its rule is (a_n=3n-2), and (3n-2=16) gives (n=6). In exams, form the general term first to find the term number.

The general term is \(a_n=n+4\). Substituting \(n=7\), we get \(a_7=7+4=11\). Therefore, 11 is the correct option. The value 10 would result if 3 were added instead of 4. Exam tip: To find a particular term, substitute its term number for \(n\) in the general rule.

When (n=1), the value should be (5), so (a_n=n+4). In exams, relate (n) to the first term in consecutive-number sequences.

\(a_n=2n-1\) generates the odd natural numbers. Checking \(n=1\) gives \(a_1=1\), and each next term increases by 2. \(2n+1\) would start with 3 instead. In exams, verify a rule using the first term.

This is an arithmetic sequence with first term 12 and common difference 12. For \(n=1\), \(a_n=12n=12\); for \(n=2\), it gives 24; and for \(n=3\), it gives 36. Hence, the correct explicit rule is \(a_n=12n\). The rule \(a_n=24n\) would give 24 as the first term, so it is incorrect. Exam tip: substitute \(n=1\) and \(n=2\) to check an explicit rule.

In option A, \(a_n\) is written only in terms of n, so any term can be found directly. B, C and D require a previous term, so they are recursive rules. Exam tip: a rule involving only n is an explicit rule.

The relevant concept is the general term of an arithmetic sequence. The consecutive difference is constant: 14 − 9 = 5, 19 − 14 = 5, and 24 − 19 = 5. The arithmetic formula is a_n = a_1 + (n − 1)d. With a_1 = 9 and d = 5, it becomes a_n = 9 + 5(n − 1) = 9 + 5n − 5 = 5n + 4. Thus option A is correct. Checking n = 1, 2, 3, and 4 gives 9, 14, 19, and 24. Option B produces 9, 18, 27, so its difference is 9. Option C starts at 1, and option D produces 9, 10, 11, so its difference is only 1. Only A fits every term.

Substitute \(n=4\): \(a_4=\frac{4(4+1)}{2}=\frac{4\times5}{2}=10\). Therefore, the correct answer is 10. Getting 12 usually results from an error in addition or multiplication. Exam tip: for a general term, substitute the given subscript for \(n\) and simplify step by step.

The governing idea is recognition of triangular numbers and their explicit formula. Each term can be expressed as the product of two consecutive positive integers divided by 2: 1 = 1×2/2, 3 = 2×3/2, 6 = 3×4/2, and 10 = 4×5/2. Therefore the term in position n is a_n = n(n + 1)/2, making option C correct. The successive differences are 2, 3, and 4, so this is not an arithmetic sequence with a constant difference. Option A gives square numbers 1, 4, 9, 16. Option B gives odd numbers 1, 3, 5, 7, while option D gives 3, 4, 5, 6. None of these alternatives reproduces all the displayed terms.

Given \(a_n=100-5n\), substitute \(n=6\): \(a_6=100-5(6)=100-30=70\). Therefore, 70 is the correct option. The value 65 may result from subtracting 5 only once, but here \(5\times6\) must be subtracted. Exam tip: To find a term from an explicit rule, substitute the given value of \(n\) and perform multiplication first.

Direct answer: B, \\(a_n=4n^2\\). The terms are 4, 16, 36 and 64. These can be written as \\(4\times1^2\\), \\(4\times2^2\\), \\(4\times3^2\\) and \\(4\times4^2\\). Therefore the rule is \\(a_n=4n^2\\). Step 1: identify the square numbers 1, 4, 9 and 16. Step 2: multiply each by 4 to obtain 4, 16, 36 and 64. Step 3: replace the position number by n. Option B gives exactly these terms. Option A gives 2, 8 and 18, so it is too small and has the wrong multiplier. Option C gives 5 at \\(n=1\\), not 4, and does not preserve the pattern. Option D gives 4, 8 and 12, which is a linear sequence rather than the given square-based sequence. Exam cue: test a proposed rule at n=1, 2 and 3 before accepting it.

Given \(a_n=2^n-1\), substitute \(n=4\): \(a_4=2^4-1=16-1=15\). Therefore, 15 is the correct option. The closest distractor, 16, is only the value of \(2^4\); the subtraction of 1 must still be done. Exam tip: substitute the term number first, then follow the order of operations.

The first term is (20) and the difference is (-3), so (a_n=23-3n). In exams, treat the common difference as negative in decreasing sequences.