Explicit or general rule

\(a_n=5n-2\) 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 direct answer is option B: a_n=4n+9. The sequence increases by 4 each time: 17-13=4, 21-17=4, and 25-21=4. For an arithmetic sequence, start with the first term and add the common difference for each step. A convenient formula is first term plus (n-1) times the difference: a_n=13+(n-1)4=13+4n-4=4n+9. Check n=1: 13; n=2: 17; n=3: 21; n=4: 25. Option A, 13n, gives 13,26,39,52, so it does not preserve the difference. Option B matches every term. Option C, 4n-9, gives -5,-1,3,7, so the starting value is wrong. Option D, n+12, increases only by 1 and gives 13,14,15,16. Memory cue: for an arithmetic sequence use a_n=a_1+(n-1)d, not a_1n.

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.

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.

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 \(n\).

The direct answer is option C: the seventh term. The sequence has a constant difference of 7: 6-(-1)=7, 13-6=7, and 20-13=7. Its arithmetic-rule form is a_n=-1+(n-1)7=-1+7n-7=7n-8. Set the required value equal to 41: 7n-8=41. Add 8: 7n=49. Divide by 7: n=7. Check: a_7=7(7)-8=49-8=41. Option A, fifth term, gives 7(5)-8=27. Option B, sixth term, gives 34. Option C, seventh term, gives 41 and is correct. Option D, eighth term, gives 48. The key is to find the position, not merely continue the list without a rule. Memory cue: equal differences indicate an arithmetic sequence; write its general term and then solve for n.

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 governing concept is substituting a specified index into an explicit sequence rule and solving the resulting linear equation. Since a_8 = 71, replace n by 8 in a_n = pn + 7. This gives 8p + 7 = 71. Subtract 7 from both sides to obtain 8p = 64, then divide by 8: p = 8. A direct check gives a_8 = 8(8) + 7 = 64 + 7 = 71, confirming the result. Therefore, option C is correct. If p = 6, the eighth term would be 55; if p = 7, it would be 63; and if p = 9, it would be 79. Thus the other choices do not satisfy the given condition. The subscript 8 is the index and must not be confused with p.

The governing concept is evaluating two terms from an explicit linear rule and then finding their difference. For n = 9, a_9 = 13(9) - 8 = 117 - 8 = 109. For n = 4, a_4 = 13(4) - 8 = 52 - 8 = 44. Therefore, a_9 - a_4 = 109 - 44 = 65, so option B is correct. There is also a useful shortcut: because the coefficient of n is 13, increasing the index from 4 to 9 by 5 increases the term by 5 × 13 = 65; the constant -8 cancels in the subtraction. Option A corresponds to only four index steps, while C and D overcount the change. Both direct calculation and the shortcut confirm 65.

The governing concept is an explicit or general rule: each term is found by substituting its positive position number into the formula. For the first term, n = 1, so a_1 = 3(1) + 4 = 7. For the second term, n = 2, so a_2 = 3(2) + 4 = 10. For the third term, n = 3, so a_3 = 3(3) + 4 = 13. Hence the first three terms are 7, 10, 13, and option B is correct. Option A begins with 4 because it incorrectly uses n = 0. Options C and D result from incorrect substitution or arithmetic. The position numbering begins at 1 for the first term unless the question explicitly states otherwise.

For the general term of an arithmetic progression, use a_n = a_1 + (n − 1)d. The first term is a_1 = 10 and the common difference is d = 15 − 10 = 5. Substitution gives a_n = 10 + (n − 1)5 = 10 + 5n − 5 = 5n + 5. Therefore option A is correct. Testing the formula confirms it: n = 1 gives 10, n = 2 gives 15, and n = 3 gives 20, exactly matching the progression. Option B gives a first term of 10 but then has difference 10. Option C gives a first term of 0, and option D has common difference 1. Thus only option A satisfies both the first term and the constant difference.

The governing concept is evaluating an explicit or general rule at successive positive integer positions. For the first three terms, use n = 1, 2, and 3 in a_n = 4n + 6. When n = 1, a_1 = 4(1) + 6 = 10; when n = 2, a_2 = 4(2) + 6 = 14; and when n = 3, a_3 = 4(3) + 6 = 18. Therefore, option B, 10, 14, 18, is correct. Option A begins with the constant 6 as though n were zero. Options C and D result from using an incorrect constant or starting value. The consecutive terms also have common difference 4, which provides an additional check.