Explicit or general rule

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.

Using (2n^2+1) gives (3,9,19,33). In exams, match the square-based rule with the first four terms.

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 first term is (2) and the difference is (5), so (a_n=5n-3). In exams, always check the first term by putting (n=1).

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.

The difference is (6) and the first term is (14), so (a_n=6n+8). In exams, take the difference as the coefficient of (n).

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 \(pn+q\).