Explicit or general rule

Substituting \(n=1,2,3,4\) in \(a_n=4n^2-1\) gives \(3,15,35,63\), respectively. Hence, the correct general term is \(4n^2-1\). Although \(a_n=3n^2\) gives the first term as 3, its second term is 12, not 15. Exam tip: verify a proposed general term by checking at least the first three values of \(n\).

The explicit rule of an AP is \(a_n=a_1+(n-1)d\). Here, \(4+3(n-1)=3n+1\), so option A is correct. Option B has common difference 4. Exam tip: substitute \(n=1\) to check the first term.

Direct answer: B, \\(a_n=3n+7\\). The sequence increases by 3 each time: 13-10=3, 16-13=3 and 19-16=3. For position n, start with the first term and add the common difference \\(n-1\\) times: \\(a_n=10+(n-1)3=10+3n-3=3n+7\\). Checking gives 10 when \\(n=1\\), 13 when \\(n=2\\), and 16 when \\(n=3\\). Option A gives 10, 20, 30, so it does not follow the observed difference. Option B gives all the required terms and is correct. Option C gives -4 at \\(n=1\\), so even the first term is wrong. Option D gives 10, 11, 12, whose difference is 1 rather than 3. Memory cue: in an arithmetic rule, the coefficient of n is the common difference; then adjust the constant using the first term.

Given \(a_n=2^{n+1}\), substitute \(n=3\): \(a_3=2^{3+1}=2^4=16\). Hence, \(16\) is correct. \(8\) would result from using \(2^3\) and incorrectly ignoring the \(+1\) in the exponent. Exam tip: substitute the term number into the entire exponent before simplifying.

Each term is twice the preceding term, so this is a geometric sequence. For \(n=1\), \(a_n=2^{n+1}=2^2=4\); for \(n=2\), it gives 8, and for \(n=3\), it gives 16. Hence, the correct general term is \(a_n=2^{n+1}\). Although \(a_n=4n\) gives the first two terms as 4 and 8, its third term is 12, not 16. Exam tip: test a proposed rule using the first two or three terms.

Substitute n=4: a_4=4^2+2(4)=16+8=24. Therefore, 24 is the correct answer. The value 20 results from incorrectly using n instead of 2n in the second term. Exam tip: when finding a term, substitute the term number into every part of the expression.

Substituting \(n=1,2,3,4\) into \(a_n=n^2+2n\) gives \(3,8,15,24\), respectively. Therefore, it is the correct general term. The close distractor \(a_n=n^2+1\) gives \(2\) when \(n=1\), so it is incorrect. Exam tip: verify a general term using at least the first three terms.

Given \(a_n=50-5n\), substitute \(n=7\) to find the seventh term: \(a_7=50-5(7)=50-35=15\). Therefore, 15 is correct. Getting 20 usually indicates an error in calculating \(5\times7\) or in subtraction. Exam tip: For an explicit rule, substitute the term number for \(n\) and simplify step by step.

At (n=1), it gives (45), and at (n=2), it gives (40), so (a_n=50-5n). In exams, always check (n=1) when forming a decreasing rule.

To find the term equal to 31, set \(6n-5=31\). This gives \(6n=36\), so \(n=6\). Hence, 31 is the sixth term of the sequence. Substituting \(n=5\) gives 25, so it is a close but incorrect option. Exam tip: when asked for a term number, equate the required value to the general term and solve for \(n\).

Its rule is (a_n=6n-5), and (6n-5=31) gives (n=6). In exams, form the general term to find the term number.

Given \(a_n=\frac{n+1}{2}\). To find \(a_5\), substitute \(n=5\): \(a_5=\frac{5+1}{2}=\frac{6}{2}=3\). Therefore, 3 is the correct option. Choosing 2 may result from incorrectly adding \(n+1\). Exam tip: substitute the term number carefully into the general-term formula.

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

In \(a_n=f(n)\), \(n\) gives the term’s position. Substituting \(n=1\) gives the first term, while \(n=2\) gives the second term. The common difference is a separate property. Exam tip: treat \(n\) as the term number.

The first term is (15) and the difference is (10), so (a_n=10n+5). In exams, use the difference as the coefficient and find the constant from the first term.

In \(a_n=5n-2\), the difference of consecutive terms is \(a_{n+1}-a_n=5\), which is constant. Hence it is an arithmetic progression, not a geometric one. Exam tip: for a linear rule \(an+b\), \(a\) is the common difference.

The direct answer is option A: a_n=2n^2+1. A general term must reproduce the sequence when n=1,2,3,4. Test option A step by step: for n=1, 2(1)^2+1=3; for n=2, 2(2)^2+1=9; for n=3, 2(3)^2+1=19; for n=4, 2(4)^2+1=33. These are exactly the given terms. Option B, 3n, gives 3,6,9,12, so the second term already fails. Option C, n^2+2, gives 3,6,11,18, so it does not match. Option D, 2^n+1, gives 3,5,9,17, which also fails. The increasing differences are 6, 10, and 14; they increase by 4, which is consistent with a quadratic rule having coefficient 2. Memory cue: substitute the first few positions into every proposed formula; one mismatch makes an option wrong.

Substitute n=3 in the given rule: a_3=3^3-2. Since 3^3=27, a_3=27-2=25. Option 27 is only the value of 3^3; the subtraction of 2 has not yet been done. Exam tip: substitute the term number first, evaluate the exponent, and then perform the remaining operation.

The sequence follows powers of 3 with 2 subtracted from each term. Evaluating the proposed rule aₙ = 3ⁿ − 2 gives, for n = 1, 3 − 2 = 1; for n = 2, 9 − 2 = 7; for n = 3, 27 − 2 = 25; and for n = 4, 81 − 2 = 79. Hence option B is correct. Option A gives 3, 9, 27, 81 because it omits the subtraction of 2. Option C gives 1, 3, 7, 15, and option D gives 0, 7, 26, 63, so neither agrees with the sequence. The key concept is an explicit exponential rule: the position determines the exponent, and the fixed adjustment is then applied.

Substitute n=6: a_6=6^2-2(6)=36-12=24. Therefore, 24 is correct. The value 22 would result from incorrectly subtracting 14 from 36. Exam tip: To find a specific term, substitute its index for n and evaluate the square first.

To identify the explicit rule, substitute the position n into each candidate and compare the resulting values. For option A, aₙ = n² − 2n: at n = 1 it gives 1 − 2 = −1; at n = 2 it gives 4 − 4 = 0; at n = 3 it gives 9 − 6 = 3; and at n = 4 it gives 16 − 8 = 8. Therefore option A exactly reproduces the sequence. Option B gives 0 as the first term, option C gives −1, 1, 3, 5, and option D gives 2, 6, 12, 20. The governing idea is that an explicit rule must produce the correct term directly for every allowed position, not merely match one or two values.

The general term is \(a_n=18+n\). Substituting \(n=5\), we get \(a_5=18+5=23\), so option C is correct. The value \(22\) would be obtained for \(n=4\), making it a close but incorrect distractor. Exam tip: To find a particular term, replace \(n\) in the general rule with that term number.

Every term is a consecutive multiple of 13: 13 = 13×1, 26 = 13×2, 39 = 13×3, and 52 = 13×4. Therefore the term in position n is aₙ = 13n, so option A is correct. The same conclusion follows from the arithmetic-sequence formula. The first term is 13 and the common difference is also 13, so aₙ = 13 + (n − 1)13 = 13n. Option B gives 14 for the first term, option C gives 13 initially but then 25 rather than 26, and option D gives 26 for the first term. Checking more than one position prevents a formula from being accepted merely because it matches the first term.

In \(a_n=5n-2\), the coefficient of \(n\) is 5, so consecutive terms have the constant difference 5. Hence it is an arithmetic progression, not a GP, which needs a constant ratio. Exam tip: a rule of the form \(an+b\) usually represents an AP.

The direct answer is option B, \(a_n=5n-3\). This is an arithmetic sequence because the same number is added each time: \(7-2=5\), \(12-7=5\), and \(17-12=5\). For an arithmetic sequence, \(a_n=a_1+(n-1)d\). Here \(a_1=2\) and \(d=5\), so \(a_n=2+(n-1)5=2+5n-5=5n-3\). Checking gives \(a_1=2\), \(a_2=7\), \(a_3=12\), and \(a_4=17\). Option A gives \(a_1=7\), so it starts incorrectly. Option B gives every listed term and is correct. Option C gives \(a_1=7\) and has difference 2, so it is wrong. Option D gives \(a_1=2\), but its difference is 7, producing 2, 9, 16, 23 rather than the given sequence. Memory cue: find the common difference and always test the formula at \(n=1\).