Explicit or general rule

Direct answer: a_n = 4n, so option A is correct. A general term gives the value of the term in position n without listing every earlier term. Here the sequence is 4, 8, 12, 16, and every term is a consecutive multiple of 4: 4×1, 4×2, 4×3, and 4×4. Therefore the nth term is 4×n, or a_n = 4n. The arithmetic-progression formula confirms this: the first term is 4 and the common difference is 4, so a_n = a₁ + (n−1)d = 4 + 4(n−1) = 4n. Option A gives 4, 8, 12, and 16 when n = 1, 2, 3, and 4. Option B has common difference 1 and begins with 5. Option C has common difference 3 and begins with 4 but soon gives 7, 10, 13. Option D begins with 3, not 4. A useful check is to test n = 1 and also compare the common difference.

In \(a_n=5n-3\), the difference \(a_{n+1}-a_n=5\) is constant, so it is an arithmetic progression. A geometric progression requires a constant ratio, not a constant difference. Exam tip: test consecutive-term differences first.

An explicit rule gives the value of any term directly from its position n. In this sequence, the first term is 11, the second is 22, the third is 33, and the fourth is 44. Each term is 11 times its position: 11 × 1 = 11, 11 × 2 = 22, 11 × 3 = 33, and 11 × 4 = 44. Therefore the general rule is aₙ = 11n, so option C is correct. Option A gives 11 for n = 1 but then gives 21 for n = 2, while option B gives 12 for the first term. Option D gives 22 as the first term, so it does not match the sequence.

\(a_n=3n-2\) has the linear form \(pn+q\). Here, \(a_{n+1}-a_n=3\), which is constant for every term, so it is an arithmetic progression. For \(n^2+1\), the differences are not constant. Exam tip: test consecutive-term differences.

Direct answer: aₙ = 5n − 2, so option A is correct. First find the common difference: 8−3 = 5, 13−8 = 5, and 18−13 = 5. Since the difference is constant, this is an arithmetic sequence. For an arithmetic sequence, aₙ = a₁ + (n−1)d. Substituting a₁ = 3 and d = 5 gives aₙ = 3 + 5(n−1) = 3 + 5n − 5 = 5n − 2. Checking n = 1 gives 3, n = 2 gives 8, n = 3 gives 13, and n = 4 gives 18. Option B begins at 8 because 3(1)+5 = 8, so it does not represent the displayed first term. Option C begins at 7 and has the wrong constant. Option D begins at 3 but has common difference 8, not 5. Thus only A matches both the first term and the repeated increase. Common warning: knowing the difference alone is not enough; the constant must also make the first term correct.

Given \(a_n=n^2+3\), substitute \(n=5\) to find the fifth term: \(a_5=5^2+3=25+3=28\). Therefore, the correct answer is \(28\). \(25\) is only the value of \(5^2\); the \(+3\) must also be added. Exam tip: Substitute the term number first, then evaluate the exponent before completing the calculation.

For \(a_n=n^2+3\), substituting \(n=1,2,3,4\) gives \(4,7,12,19\), respectively. Hence, \(a_n=n^2+3\) is the correct general term. Although \(a_n=3n+1\) gives the first two terms 4 and 7, its third term is 10, not 12. Exam tip: test a proposed general term using at least the first three values of \(n\).

Using the rule \(a_n=14-3n\), substitute \(n=3\): \(a_3=14-3(3)=14-9=5\). Therefore, 5 is correct. Getting 7 would result from subtracting only 3 from 14 instead of calculating \(3\times3\). Exam tip: substitute the term number first, then perform multiplication before subtraction.

At (n=1) it gives (11), and at (n=2) it gives (8), so (a_n=14-3n). In exams, treat a decreasing difference as negative.

The general term is \(a_n=5^n\). Substituting \(n=2\) gives \(a_2=5^2=25\). The nearby distractor \(10\) is incorrect because it is not the square of \(5\). Exam tip: To find a particular term from a general rule, substitute the term number for \(n\).

The first term is 5, and each successive term is multiplied by 5: \(5,5^2,5^3,5^4,\ldots\). Therefore, when counting starts at \(n=1\), the general term is \(a_n=5^n\). The rule \(a_n=5^{n-1}\) gives 1 as the first term, so it is incorrect. Exam tip: substitute \(n=1\) to check a proposed general rule.

It is an arithmetic progression because \(a_{n+1}-a_n=[4(n+1)+7]-(4n+7)=4\), a constant difference. A geometric progression needs a constant ratio, not a difference. Exam tip: in \(a_n=dn+c\), \(d\) is the common difference.

Substituting \(n=1,2,3,4\) in \(a_n=3n^2\) gives \(3,12,27,48\), respectively. Therefore, the correct general term is \(a_n=3n^2\). Although \(a_n=3n\) looks related, it gives \(3,6,9,12\), so it is not correct. Exam tip: verify a proposed general term using at least the first three terms.

Given \(a_n=\frac{3n}{2}\). To find the fourth term, substitute \(n=4\): \(a_4=\frac{3\times4}{2}=\frac{12}{2}=6\). Therefore, the correct answer is 6. A value such as 5 may result from substituting an incorrect value of \(n\). Exam tip: to find a particular term, substitute its term number in the general-term formula.

The direct answer is option C: the ninth term. The rule says the term at position n is a_n=2n-3, and we need its value to be 15. Therefore set 2n-3=15. Add 3 to both sides: 2n=18. Divide by 2: n=9. Check: a_9=2(9)-3=18-3=15. Option A, n=7, gives 14-3=11, not 15. Option B, n=8, gives 16-3=13, not 15. Option C, n=9, gives 15, so it is correct. Option D, n=10, gives 20-3=17, not 15. The common mistake is to stop at 2n=18 or to confuse the term value 15 with the term number. Exam cue: when a term value is given, put it equal to the formula and solve for n.

Its rule is (a_n=2n-3), and (2n-3=15) gives (n=9). In exams, form the general term first to find the term number.

Given \(a_n=n^3\). For the third term, substitute \(n=3\): \(a_3=3^3=3\times3\times3=27\). Option 9 is \(3^2\), so it would result from squaring rather than cubing. Exam tip: in an explicit rule, substitute the term number directly for \(n\).

Substitute 2 for n in the general term: \(a_2=2^3+1=8+1=9\). Therefore, the correct answer is 9. Option 8 is only the value of \(2^3\); it misses the given \(+1\). Exam tip: while finding a term of a sequence, substitute the value of n first and then follow the order of operations.

For \(a_n=n^3+1\), substituting \(n=1,2,3,4\) gives \(2,9,28,65\), respectively. Hence, option B is correct. In option A, the third term would be \(10\), not \(28\). Exam tip: verify a proposed general term by checking at least the first three values of \(n\).

Given \(a_n=20-4n\). To find the fourth term, substitute \(n=4\): \(a_4=20-4(4)=20-16=4\). Therefore, 4 is the correct answer. Option 8 could result from incorrectly subtracting 12 instead of calculating \(4\times4\). Exam tip: In an explicit rule, substitute the term number carefully for \(n\).

At (n=1) it gives (16), and at (n=2) it gives (12), so (a_n=20-4n). In exams, check the rule with the first term in a decreasing sequence.

In \(a_n=4n-3\), the highest power of \(n\) is 1, so it is a linear general term and consecutive terms have a constant difference. \(n^2+1\) is quadratic. Exam tip: check whether the highest power of \(n\) is 1.

Given \(a_n=\frac{n(n-1)}{2}\), substitute \(n=5\): \(a_5=\frac{5(5-1)}{2}=\frac{5\times4}{2}=10\). Hence, 10 is the correct option. The value 15 would result from using \(\frac{n(n+1)}{2}\), which is not the given rule. Exam tip: to find a particular term from a general rule, first substitute its subscript for \(n\).

Putting (n=1,2,3,4) gives (0,1,3,6). In exams, test triangular-number-like options with small (n).

Option A squares the position number \(n\), so its first three terms are 1, 4 and 9. In contrast, \(2n-1\) generates odd numbers, not square numbers. Exam tip: identify the operation applied to \(n\) before listing terms.