Explicit or general rule

Option A is correct because substituting \(n=1\) gives the first term 3, and each successive term increases by 5. Option D also has common difference 5, but its first term is 2. In exams, check both first term and difference.

Direct answer: A, \\(a_n=6n+8\\). The common difference is 6 because 20-14=6, 26-20=6 and 32-26=6. Use the arithmetic-sequence formula: \\(a_n=a_1+(n-1)d\\). Substituting \\(a_1=14\\) and \\(d=6\\) gives \\(a_n=14+(n-1)6=14+6n-6=6n+8\\). For \\(n=1\\), it gives 14; for \\(n=2\\), 20; and for \\(n=3\\), 26. Option A is therefore correct. Option B gives 14, 28, 42 and wrongly uses 14 as the difference. Option C gives 6 at \\(n=1\\), not 14. Option D gives 14, 22, 30, so its difference is 8. Exam cue: first find the common difference; it becomes the coefficient of n in an arithmetic sequence.

For the fourth term, substitute n=4 in the rule: (a_4=4^2+4=16+4=20). Therefore, the correct answer is 20. The value 16 is only the square of 4; the constant 4 in the rule still has to be added. Exam tip: In an explicit rule, substitute the term number for the variable and follow the order of operations carefully.

The correct general term is \(a_n=n^2+4\). Substituting \(n=1,2,3,4\) gives \(5,8,13,20\), respectively. The closest option, \(a_n=n^2+3\), gives \(4\) as the first term when \(n=1\), so it is incorrect. Exam tip: test a proposed general term by substituting the first two or three values of \(n\).

Given \(a_n=30-2n\), substitute \(n=7\) to find the seventh term: \(a_7=30-2(7)=30-14=16\). Therefore, 16 is correct. The value 14 would result from using an incorrect term number or calculation. Exam tip: substitute the required term number carefully for \(n\) in the general term.

At (n=1) it gives (28), and at (n=2) it gives (26), so (a_n=30-2n). In exams, treat a decreasing difference as negative.

The general rule is (a_n=4^n). Substituting n=2 gives (a_2=4^2=4×4=16). Hence, 8 is incorrect because it is not the square of 4. Exam tip: In an explicit rule, substitute the required term number for n.

The sequence is formed by consecutive powers of 4. With the first position numbered n = 1, the terms are 4¹ = 4, 4² = 16, 4³ = 64, and 4⁴ = 256. Hence the explicit rule is aₙ = 4ⁿ, making option C correct. Equivalently, each term is obtained by multiplying the previous term by 4, which is the characteristic pattern of a geometric sequence with first term 4 and common ratio 4. Option A gives 4, 8, 12, 16, while option B gives 1, 16, 81, 256. Option D gives 3, 15, 63, 255. Only option C agrees with every listed position.

For \(a_n=3n+1\), increasing \(n\) by 1 increases every term by 3, so the common difference is constant. In \(n^2+1\), differences change. Exam tip: identify AP rules of the form \(pn+q\).

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.

The direct answer is option C: n=8. We want the term value to be 25, while the rule is a_n=4n-7. Set 4n-7=25. Add 7: 4n=32. Divide by 4: n=8. Verification gives a_8=4(8)-7=32-7=25. Option A, n=6, gives 24-7=17, not 25. Option B, n=7, gives 28-7=21, not 25. Option C, n=8, gives exactly 25, so it is correct. Option D, n=9, gives 36-7=29, not 25. The term number n tells the position, whereas 25 is the value at that position. A useful exam method is to isolate n carefully: undo subtraction first, then undo multiplication. Do not divide before adding 7, because that would change the equation incorrectly.

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 \(3n\).

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.

The direct answer is option B: a_n=\frac{n(n+3)}{2}. To test a general term, use n=1,2,3,4 because these positions produce the listed terms. For option B: n=1 gives 1(4)/2=2; n=2 gives 2(5)/2=5; n=3 gives 3(6)/2=9; n=4 gives 4(7)/2=14. Thus it gives 2,5,9,14 exactly. Option A, n+1, gives 2,3,4,5 and misses from the second term. Option C, n^2+1, gives 2,5,10,17, so the third term is wrong. Option D, 2n+1, gives 3,5,7,9 and fails at the first term. The successive differences are 3,4,5, so the rule is not a simple constant-difference linear sequence. Memory cue: for a formula with a fraction, substitute small positions and simplify completely before comparing.

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.

The direct answer is option A: a_n=6n^2-2. Substitute n=1,2,3,4. Option A gives 6(1)^2-2=4, 6(2)^2-2=24-2=22, 6(3)^2-2=54-2=52, and 6(4)^2-2=96-2=94. These exactly reproduce the sequence. Option B, 4n, gives 4,8,12,16, so only the first term agrees. Option C, 6n-2, gives 4,10,16,22, so it is linear and does not match. Option D, 2n^2+2, gives 4,10,20,34, so the second term already fails. The differences in the given sequence are 18,30,42, increasing by 12, which supports a quadratic rule with coefficient 6. Exam cue: when differences themselves grow regularly, test a square term, then verify several positions rather than just one.