Explicit or general rule

This is a geometric sequence because each term is 3 times the preceding term. For \(n=1\), the first term must be \(9=3^2\). Hence the exponent is \(n+1\), so \(a_n=3^{n+1}\). The rule \(a_n=3^n\) gives 3 as the first term, not 9. Exam tip: always substitute \(n=1\) to check a proposed general term.

Substitute 4 for n: \(a_4=4^2+3(4)=16+12=28\). Hence, 28 is correct. Getting 26 usually results from an error in calculating \(3\times4\) or in addition. Exam tip: To find a term from an explicit rule, replace every n with the given term number and simplify step by step.

The terms do not have a constant first difference: the differences are 6, 8, and 10. Their second differences are constant, since 8 − 6 = 2 and 10 − 8 = 2, so a quadratic rule is appropriate. Test option A at the first values of n: when n = 1, a₁ = 1² + 3(1) = 4; when n = 2, a₂ = 2² + 3(2) = 10; when n = 3, a₃ = 9 + 9 = 18; and when n = 4, a₄ = 16 + 12 = 28. Thus option A reproduces all four terms and is correct. Option B and option D are linear, so their differences would be constant. Option C gives 3, 8, 15, 24, not the stated sequence.

When \(n\) increases by 1, \(a_n=5n-2\) increases by 5 each time, so it is an arithmetic progression with common difference 5. The \(-2\) is a constant term, not the common difference. Exam tip: the coefficient of \(n\) gives the common difference.

At (n=1) it gives (54), and at (n=2) it gives (48), so (a_n=60-6n). In exams, check a decreasing sequence rule using (n=1).

Given \(a_n=7n-8\), set the term equal to 41: \(7n-8=41\). Thus, \(7n=49\), so \(n=7\). Therefore, 41 is the seventh term. Substituting \(n=6\) gives 34, not 41. Exam tip: To find the term number, equate \(a_n\) to the given value and solve for \(n\).

Its rule is (a_n=7n-8), and (7n-8=41) gives (n=7). In exams, find the term number from the general term.

Given \(a_n=\frac{2n+1}{3}\). Substituting \(n=4\), we get \(a_4=\frac{2(4)+1}{3}=\frac{9}{3}=3\). The value \(\frac{7}{3}\) would result from using \(n=3\), so it is a close but incorrect option. Exam tip: substitute the required term number carefully for \(n\) before simplifying.

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

The correct rule is \(a_n=4n+1\), since \(a_{n+1}-a_n=[4(n+1)+1]-(4n+1)=4\). The other rules do not have a constant difference. Exam tip: compare consecutive terms’ differences.

The first term is (15) and the difference is (11), so (a_n=11n+4). In exams, use the difference as coefficient and find the constant from the first term.

\(a_n=5n-2\) has the linear form \(pn+q\). Here, \(a_{n+1}-a_n=5\), so the common difference is constant. For \(n^2+1\), consecutive differences change. Exam tip: a first power of \(n\) indicates a possible arithmetic sequence.

For \(a_n=3n^2+2n\), substituting \(n=1,2,3,4\) gives \(5,16,33,56\), respectively. Hence, it is the required general term. Although \(a_n=2n^2+3n\) gives the first term as 5, it gives 14 as the second term, not 16. Exam tip: verify a proposed general term by checking at least the first two or three values of \(n\).

Given \(a_n=2^n+3\), substitute \(n=4\): \(a_4=2^4+3=16+3=19\). Therefore, 19 is the correct option. The value 18 would result from adding 2 to \(2^4\), but the rule requires adding 3. Exam tip: substitute the term number first, then evaluate the exponent.

Here the terms are counted from \(n=1\). Substituting \(n=1,2,3,4\) in \(a_n=2^n+3\) gives \(5,7,11,19\), respectively, so it is the correct explicit rule. \(a_n=2n+3\) matches only the first two terms; for the third term it gives \(9\), not \(11\). Exam tip: test an explicit rule using at least the first three values of \(n\).

Given \(a_n=n^2-3n\). Substituting \(n=7\), \(a_7=7^2-3(7)=49-21=28\). Therefore, 28 is the correct option. A value such as 24 can result from an error while calculating \(7^2\) or \(3\times7\). Exam tip: To find a particular term, substitute the given term number carefully for \(n\) in the formula.

To identify the rule, substitute the term number n into each candidate. For option A, a₁ = 1² − 3(1) = −2, a₂ = 2² − 3(2) = −2, a₃ = 3² − 3(3) = 0, and a₄ = 4² − 3(4) = 4. These are exactly the four given terms, so aₙ = n² − 3n is correct. The rule may contain negative or zero values; that does not make it invalid. Option B produces −1, 2, 7, 14, and option C produces −2, 0, 2, 4, so neither matches the sequence. Option D gives 2, 6, 12, 20 and is also inconsistent. Testing several initial indices is a reliable way to verify an explicit sequence rule.

Given \(a_n=25+n\), substitute \(n=6\) for the sixth term: \(a_6=25+6=31\). Hence, 31 is the correct option. The value 30 would result from using \(n=5\), so it is a close but incorrect distractor. Exam tip: carefully substitute the required term number for \(n\) in the general term.

The first term is (4) and the difference is (5), so (a_n=5n-1). In exams, always check the first term by putting (n=1).

In \(a_n=5n+2\), the coefficient of \(n\) is constant, so consecutive terms always differ by 5. Hence it is an arithmetic progression. Rules involving \(n^2\) or \(2^n\) do not have a constant difference. Exam tip: a rule of the form \(an+b\) represents an AP.