Explicit or general rule

For \(a_n=5n-2\), \(a_{n+1}-a_n=[5(n+1)-2]-(5n-2)=5\), which is constant; hence it is an arithmetic progression. In \(n^2+1\), the differences vary. Exam tip: look for a linear rule \(pn+q\).

Substituting \(n=1,2,3,4\) in \(a_n=2n^2+3\) gives \(5,11,21,35\), respectively, so option C is correct. The close distractor \(a_n=3n^2+2\) gives the first term as 5, but for \(n=2\) it gives 14, not 11. Exam tip: test an explicit rule using at least the first three values of \(n\).

For \(a_n=5n-2\), \(a_{n+1}-a_n=5\) for every \(n\), so the common difference is constant. In \(n^2+1\), the differences change. Exam tip: test consecutive-term differences.

The first differences are \(8,14,20\), and the second differences are constant: \(6,6\). Hence, the general term should be quadratic. For \(a_n=3n^2-n\), substituting \(n=1,2,3,4\) gives \(2,10,24,44\), respectively. Option A gives \(8\) as the second term, while option D gives \(12\), so they are incorrect. Exam tip: test a proposed rule by substituting at least the first three values of \(n\).

Given \(a_n=n^3+n\), substitute \(n=3\): \(a_3=3^3+3=27+3=30\). Option 27 represents only \(3^3\); it misses the \(+n\) term in the rule. Exam tip: To find a term of a sequence, substitute the required index into every part of the formula.

Substituting \(n=1,2,3,4\) in \(a_n=n^3+n\) gives \(2,10,30,68\), respectively. Therefore, the correct general term is \(a_n=n^3+n\). The close distractor \(a_n=n^3+1\) gives the first term as 2, but for \(n=2\) it gives 9, not 10. Exam tip: verify a proposed general term using at least the first two or three terms.

Given \(a_n=2n-1\), put \(a_n=41\). Then \(2n-1=41\), so \(2n=42\) and hence \(n=21\). Therefore, 41 is the 21st term of the sequence. For example, if \(n=20\), then \(a_{20}=39\), not 41. Exam tip: To find the term number of a given value, substitute that value in the general term and solve for \(n\).

Its rule is (a_n=3n+4), and (3n+4=58) gives (n=18). In exams, write the rule first and compare it with the given term.

The difference is (6) and the first term is (8), so (a_n=6n+2). In exams, find the constant to match the first term.

The first three terms are (8,14,20), and their sum is (42). In exams, find all required terms when a sum is asked.

Given \(a_n=4n-1\), \(a_7=4\times7-1=27\) and \(a_3=4\times3-1=11\). Therefore, \(a_7-a_3=27-11=16\). The option 14 would result from an incorrect subtraction of the terms. Exam tip: substitute the value of \(n\) and find each required term separately.

Given \(a_n=n^2+5n\), \(a_4=4^2+5(4)=16+20=36\) and \(a_2=2^2+5(2)=4+10=14\). Therefore, \(a_4:a_2=36:14=18:7\). Option (36:7) is not correct because only the first term has effectively been simplified. Exam tip: always reduce a ratio by dividing both terms by their greatest common factor.

For \(a_n=5n-2\), increasing \(n\) by 1 increases every term by 5, so the common difference is constant. In \(n^2+1\), differences vary. Exam tip: an AP rule has the linear form \(pn+q\).

The first differences are 4, 6, and 8, so they are not constant. The second differences are 2 and 2, indicating a quadratic explicit rule. Test option C: for n = 1, a₁ = 1² + 1 − 1 = 1; for n = 2, a₂ = 4 + 2 − 1 = 5; for n = 3, a₃ = 9 + 3 − 1 = 11; and for n = 4, a₄ = 16 + 4 − 1 = 19. Therefore, option C exactly generates the sequence. Option A gives 2, 5, 10, 17; option B gives 1, 7, 17, 31; and option D gives 3, 8, 15, 24. Constant second differences are a useful clue for selecting a quadratic rule.

Given \(a_n=n^2+n-1\), substitute \(n=6\): \(a_6=6^2+6-1=36+6-1=41\). Therefore, \(41\) is correct. The distractor \(40\) may result from an error in evaluating the expression or overlooking the final subtraction. Exam tip: after substituting a value in a general term, check the square, addition, and subtraction in order.

Given \(a_n=100-7n\), substitute \(n=8\): \(a_8=100-7\times8=100-56=44\). Therefore, option B is correct. Getting 42 would result from an incorrect calculation, since \(7\times8=56\). Exam tip: after substituting the term number, perform multiplication before subtraction.

At (n=1) it gives (93), and at (n=2) it gives (86), so (a_n=100-7n). In exams, match the first term of a decreasing sequence.

To find the fifth term, substitute \(n=5\): \(a_5=\frac{3(5)+1}{2}=\frac{16}{2}=8\). Hence, 8 is the correct option. A value such as 7 can result from an incorrect calculation of the numerator \(3n+1\). Exam tip: substitute the term number directly for \(n\) in the general-term formula.

(\frac{3n+1}{2}) gives (2,\frac{7}{2},5,\frac{13}{2}). In exams, match fractional terms using small (n) values.

Given \(a_n=\frac{n(n+1)}{2}\). Substituting \(n=6\), \(a_6=\frac{6(6+1)}{2}=\frac{6\times7}{2}=21\). Therefore, 21 is correct. A value such as 18 can result from using an incorrect value for \(n+1\). Exam tip: substitute the term number carefully and evaluate the brackets before simplifying.