Explicit or general rule

The terms do not have a constant first difference: the differences are 15, 25, and 35. However, these differences increase by 10, suggesting a quadratic rule. The proposed expression \(5n^2+2\) can be checked directly by substituting the small positive indices. Testing the first few positions is an efficient way to verify a general term when answer choices are provided.

For \(n=1\), \(5(1)^2+2=7\); for \(n=2\), \(5(2)^2+2=22\); for \(n=3\), it gives 47; and for \(n=4\), it gives 82. Thus it reproduces every displayed term, so option A is correct. The linear choices cannot produce differences that grow in this way.

Given \(a_n=\frac{5n}{2}\). To find the sixth term, substitute \(n=6\): \(a_6=\frac{5\times6}{2}=\frac{30}{2}=15\). Therefore, 15 is correct. A value such as 12 can result from substituting an incorrect value for \(n\). Exam tip: carefully substitute the term number into the general-term formula.

The sequence consists of consecutive multiples of 5/2. Substituting n = 1, 2, 3, and 4 into aₙ = 5n/2 gives 5/2, 10/2 = 5, 15/2, and 20/2 = 10, respectively. Therefore option A is correct. The arithmetic-sequence formula confirms this: the first term is 5/2 and the common difference is also 5/2, so aₙ = 5/2 + (n − 1)(5/2) = 5n/2. Option B gives 1/5 for the first term, option C gives 5, and option D gives 7, so none of those matches the first term and the subsequent pattern. The fraction should remain attached to n as the multiplier 5/2.

From (4n-7=25), we get (n=8). In exams, when the term number is asked, equate the formula to the given value.

Its rule is (a_n=4n-7), and (4n-7=25) gives (n=8). In exams, form the general term first to find the term number.

Given \(a_n=n^3-1\). Substituting \(n=3\), we get \(a_3=3^3-1=27-1=26\). Therefore, 26 is correct. The close distractor 27 is only \(3^3\); the subtraction of 1 must still be done. Exam tip: For a sequence term, substitute the required index in the rule before calculating.

Taking the term number from \(n=1\), the rule \(a_n=n^3-1\) gives \(1^3-1=0\), \(2^3-1=7\), \(3^3-1=26\), and \(4^3-1=63\). Hence, option B is correct. The close distractor \(n^2-1\) gives \(3\) as its second term, not \(7\). Exam tip: substitute at least the first three values of \(n\) to verify a general term.

For \(a_n=5n-2\), the difference of consecutive terms is constant: \(a_{n+1}-a_n=[5(n+1)-2]-(5n-2)=5\). Hence it is an arithmetic progression. In \(n^2+1\), the differences change. Exam tip: a rule of the form \(pn+q\) generates an AP.

Substituting \(n=1,2,3,4\) into \(2n^3\) gives \(2,16,54,128\), respectively. Hence, the general term is \(a_n=2n^3\). The closest distractor, \(2n^2\), gives the first term as 2, but for \(n=2\) it gives 8, not 16. Exam tip: verify a general term by checking it against at least the first three terms.

Given \(a_n=40-3n\), substitute \(n=5\) to find the fifth term: \(a_5=40-3(5)=40-15=25\). Therefore, 25 is correct. A value such as 23 can result from an error in multiplication or subtraction. Exam tip: after substituting the term number, multiply before subtracting.

At (n=1) it gives (37), and at (n=2) it gives (34), so (a_n=40-3n). In exams, check a decreasing rule with the first term.

For term number \(n\), the description gives three times \(n\) minus 1, or \(3n-1\). Hence \(a_n=3n-1\). Checking with \(n=1\) gives the first term as 2. Exam tip: translate “three times” directly as \(3n\).

This is an arithmetic sequence because each term increases by the same amount: 20 − 11 = 9, 29 − 20 = 9, and 38 − 29 = 9. For an arithmetic sequence, the explicit rule is aₙ = a₁ + (n − 1)d, where a₁ is the first term and d is the common difference. Substituting a₁ = 11 and d = 9 gives aₙ = 11 + 9(n − 1) = 11 + 9n − 9 = 9n + 2. Therefore, option B is correct. Checking n = 1 gives 11, n = 2 gives 20, and n = 3 gives 29. Option A has the wrong growth, while options C and D fail even at the first term.

For the third term, substitute \(n=3\): \(a_3=\frac{3(3+3)}{2}=\frac{3\times6}{2}=9\). Therefore, the correct answer is 9. The value 12 may result from forgetting to divide by 2. Exam tip: substitute the term number carefully into the general term and simplify step by step.

(\frac{n(n+3)}{2}) gives (2,5,9,14). In exams, test fractional formulas with small (n) values.

For \(a_n=4n+1\), \(a_{n+1}-a_n=[4(n+1)+1]-(4n+1)=4\), a constant difference. In \(n^2+1\), the differences vary. Exam tip: a rule of the form \(pn+q\) is arithmetic.

Using (6n^2-2) gives (4,22,52,94). In exams, match the square rule with the first three terms.

\(a_n=5n-2\) has the form \(dn+c\), with constant \(d=5\). Hence consecutive terms differ by 5, so it is an arithmetic progression. For \(n^2+1\), the differences are not constant. Exam tip: check whether the power of \(n\) is 1.

The difference is (4) and the first term is (13), so (a_n=4n+9). In exams, match both the difference and the first term.

Given \(a_n=3^{n+1}\). Substituting \(n=2\), we get \(a_2=3^{2+1}=3^3=27\). Therefore, 27 is the correct option. Values such as 18 or 24 may result from incorrectly substituting into the exponent \(n+1\). Exam tip: while finding a term, substitute the value of \(n\) carefully into the exponent first.