Explicit or general rule

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.

Substituting \(n=1,2,3,4\) in \(a_n=4n^2-1\) gives \(3,15,35,63\), respectively. Hence, the correct general term is \(4n^2-1\). Although \(a_n=3n^2\) gives the first term as 3, its second term is 12, not 15. Exam tip: verify a proposed general term by checking at least the first three values of \(n\).

The explicit rule of an AP is \(a_n=a_1+(n-1)d\). Here, \(4+3(n-1)=3n+1\), so option A is correct. Option B has common difference 4. Exam tip: substitute \(n=1\) to check the first term.

The difference is (3) and the first term is (10), so (a_n=3n+7). In exams, check the rule on the first two terms.

Given \(a_n=2^{n+1}\), substitute \(n=3\): \(a_3=2^{3+1}=2^4=16\). Hence, \(16\) is correct. \(8\) would result from using \(2^3\) and incorrectly ignoring the \(+1\) in the exponent. Exam tip: substitute the term number into the entire exponent before simplifying.

Each term is twice the preceding term, so this is a geometric sequence. For \(n=1\), \(a_n=2^{n+1}=2^2=4\); for \(n=2\), it gives 8, and for \(n=3\), it gives 16. Hence, the correct general term is \(a_n=2^{n+1}\). Although \(a_n=4n\) gives the first two terms as 4 and 8, its third term is 12, not 16. Exam tip: test a proposed rule using the first two or three terms.

Substitute n=4: a_4=4^2+2(4)=16+8=24. Therefore, 24 is the correct answer. The value 20 results from incorrectly using n instead of 2n in the second term. Exam tip: when finding a term, substitute the term number into every part of the expression.

Substituting \(n=1,2,3,4\) into \(a_n=n^2+2n\) gives \(3,8,15,24\), respectively. Therefore, it is the correct general term. The close distractor \(a_n=n^2+1\) gives \(2\) when \(n=1\), so it is incorrect. Exam tip: verify a general term using at least the first three terms.

Given \(a_n=50-5n\), substitute \(n=7\) to find the seventh term: \(a_7=50-5(7)=50-35=15\). Therefore, 15 is correct. Getting 20 usually indicates an error in calculating \(5\times7\) or in subtraction. Exam tip: For an explicit rule, substitute the term number for \(n\) and simplify step by step.

At (n=1), it gives (45), and at (n=2), it gives (40), so (a_n=50-5n). In exams, always check (n=1) when forming a decreasing rule.

To find the term equal to 31, set \(6n-5=31\). This gives \(6n=36\), so \(n=6\). Hence, 31 is the sixth term of the sequence. Substituting \(n=5\) gives 25, so it is a close but incorrect option. Exam tip: when asked for a term number, equate the required value to the general term and solve for \(n\).

Its rule is (a_n=6n-5), and (6n-5=31) gives (n=6). In exams, form the general term to find the term number.

Given \(a_n=\frac{n+1}{2}\). To find \(a_5\), substitute \(n=5\): \(a_5=\frac{5+1}{2}=\frac{6}{2}=3\). Therefore, 3 is the correct option. Choosing 2 may result from incorrectly adding \(n+1\). Exam tip: substitute the term number carefully into the general-term formula.

(\frac{n+1}{2}) gives (1,\frac{3}{2},2,\frac{5}{2}). In exams, match fractional terms using small (n) values.

In \(a_n=f(n)\), \(n\) gives the term’s position. Substituting \(n=1\) gives the first term, while \(n=2\) gives the second term. The common difference is a separate property. Exam tip: treat \(n\) as the term number.

The first term is (15) and the difference is (10), so (a_n=10n+5). In exams, use the difference as the coefficient and find the constant from the first term.

In \(a_n=5n-2\), the difference of consecutive terms is \(a_{n+1}-a_n=5\), which is constant. Hence it is an arithmetic progression, not a geometric one. Exam tip: for a linear rule \(an+b\), \(a\) is the common difference.