Explicit or general rule

Given \(a_n=4^n-2\), substitute \(n=3\): \(a_3=4^3-2=64-2=62\). Hence, \(62\) is correct. The close distractor \(64\) is only the value of \(4^3\); the subtraction of \(2\) must still be done. Exam tip: substitute the term number first, then evaluate the exponent before subtracting.

The numerator is (3n+1) and the denominator is (2n+3), so (a_n=\frac{3n+1}{2n+3}). In a fractional sequence form rules for both parts separately.

The direct answer is option B: \(a_n=4n\). An arithmetic sequence has the same difference between consecutive terms. The third term is 12 and the seventh term is 28. From the third term to the seventh term there are four equal steps: from \(a_3\) to \(a_4\), then \(a_5\), \(a_6\), and \(a_7\). Therefore, the total increase is \(28-12=16\), so the common difference is \(16\div4=4\). Using \(a_n=a_1+(n-1)d\), or simply testing the rule, \(4n\) gives \(a_3=4(3)=12\) and \(a_7=4(7)=28\), so it fits both facts. Option A, \(3n+3\), gives \(a_3=12\) but \(a_7=24\), so it fails. Option B gives both given terms correctly and is therefore correct. Option C, \(4n+1\), gives 13 and 29, not 12 and 28. Option D, \(5n-3\), gives \(a_3=12\) but \(a_7=32\), so it also fails. A useful exam check is to substitute both known term numbers, not just one. Remember: the number of gaps between the third and seventh terms is \(7-3=4\), not seven.

For the first three terms, substitute \(n=1,2,3\). Thus, \(a_1=2(1)^2-1+4=5\), \(a_2=2(2)^2-2+4=10\), and \(a_3=2(3)^2-3+4=19\). Therefore, the correct sequence is \((5, 10, 19)\). Option C cannot be correct because its first term is 6. Exam tip: when a general term is given, begin with \(n=1\) and calculate each term systematically.

Given (a_n=mn-4), put n=6 to get (a_6=6m-4). Thus, (6m-4=50), so (6m=54) and (m=9). Hence, 9 is the correct option. For example, if m=8, then (a_6=44), not 50. Exam tip: When a particular term is given, substitute its index in the general-term rule.

The first term is (9) and the difference is (-7), so (a_n=9+(n-1)(-7)=16-7n). When forming a rule from words, treat decrease as a negative difference.

The terms are (1\cdot2,2\cdot3,3\cdot4,4\cdot5), so (a_n=n(n+1)). Identify the product of two consecutive natural numbers.

The correct rule is \(a_n=n^2+4n+1\). Substituting \(n=1,2,3,4\) gives \(6,13,22,33\), respectively. The first differences are \(7,9,11\), so the constant second difference is \(2\), indicating a quadratic rule. In option B, the second term is \(14\), not the given \(13\). Exam tip: verify a general-term rule by substituting \(n=1,2,3\) and checking the first few terms.

From (\frac{2n}{n+3}=\frac{4}{5}), (10n=4n+12) and (n=2), so none of the given options is correct. Use cross multiplication in a fractional equation.

Given \(a_n=3n^2+4\), \(a_2=3(2)^2+4=16\) and \(a_5=3(5)^2+4=79\). Therefore, \(a_2+a_5=16+79=95\). A value such as \(93\) usually results from an error while squaring or adding a term. Exam tip: evaluate each required term separately before adding them.

The terms \(9,25,49,81\) are \(3^2,5^2,7^2,9^2\), respectively. They are squares of consecutive odd numbers, represented by \(2n+1\). For \(n=1\), \((2\times1+1)^2=9\). Hence, the general term is \(a_n=(2n+1)^2\). Option B gives \(1\) as its first term, so it does not represent this sequence. Exam tip: verify a general term by substituting \(n=1\) and \(n=2\).

Substituting \(n=1,2,3,4\) into \(a_n=n^2+2\) gives \(3,6,11,18\), respectively. The successive differences are \(3,5,7\), which are odd numbers, indicating a square-based sequence. In contrast, \(a_n=n^2+1\) gives the first term as \(2\), so it is not correct. Exam tip: when second differences are constant or first differences are consecutive odd numbers, test a rule involving \(n^2\).

Given \(a_n=3^n+2n\): for \(n=1\), \(a_1=3^1+2(1)=5\); for \(n=2\), \(a_2=3^2+2(2)=13\); and for \(n=3\), \(a_3=3^3+2(3)=33\). Therefore, the first three terms are \((5, 13, 33)\). Option C has an incorrect second term because \(3^2+4=13\), not 14. Exam tip: evaluate and add both parts, \(3^n\) and \(2n\), for every value of \(n\).

The total change over five gaps is (-15), so (d=-3), hence (a_n=14+(n-1)(-3)=17-3n). Find the common difference first.

Given n² + 4n = 77, we get n² + 4n − 77 = 0, or (n − 7)(n + 11) = 0. Thus, n = 7 or n = −11. Since a sequence index must be a positive integer, n = 7 is valid. Although −11 is an algebraic root, it cannot represent a term position. Exam tip: after solving, always check whether the value is an allowed sequence index.

Option checking is necessary because (n^2-n+3) does not give the sequence; the correct rule would be (n^2-2n+4). This type tests consistency of options.

The governing concept is direct substitution into an explicit fractional rule. To find a₃, replace every occurrence of n by 3, not just the n in the numerator. Thus a₃ = [3(3) + 2]/[2(3) − 1] = (9 + 2)/(6 − 1) = 11/5. Therefore option C is correct. Option A results from reducing the numerator incorrectly, option B omits the added 2 in the numerator, and option D uses an incorrect numerator value. The denominator must also be evaluated carefully; it is 5, not 6, because the expression is 2n − 1. Since 11 and 5 have no common factor, 11/5 is already in simplest form.

The direct answer is B: \\(a_n=(n+4)^3\\). Take cube roots of the terms: \\(125=5^3\\), \\(216=6^3\\), \\(343=7^3\\), and \\(512=8^3\\). The cube bases are 5, 6, 7, 8, so for the nth term the base is \\(n+4\\): when n=1 it is 5, and when n=4 it is 8. Cubing gives \\(a_n=(n+4)^3\\). Option B is correct. Option A, \\((n+3)^3\\), starts with \\(4^3=64\\), not 125. Option C, \\(n^3+124\\), gives the first term 125 but the second term 132, not 216; matching one term is not enough. Option D, \\(5n^3\\), gives the first term 5, not 125. Substitution confirms option B: at n=1, \\(5^3=125\\), and at n=2, \\(6^3=216\\). Memory cue: for cube sequences, first identify the cube roots, then find the rule for those roots.

In option A, substituting \(n=1,2,3,4\) gives \(9,20,33,48\), respectively. Therefore, the correct rule is \(a_n=n^2+8n\). Option B gives the first term as 9, but for \(n=2\) it gives 19, not 20. Exam tip: Test a proposed rule by substituting values of \(n\) for at least the first three terms.

The direct answer is option A: \(-4\). The rule is \(a_n=5-4n\). In a linear formula of the form \(a_n=c+dn\), the coefficient of \(n\) is the common difference \(d\). We can also prove this by finding consecutive terms. The first term is \(a_1=5-4(1)=1\), the second is \(a_2=5-4(2)=-3\), and the difference is \(-3-1=-4\). The third term is \(-7\), and again \(-7-(-3)=-4\). Thus every step decreases by 4. Option A, \(-4\), is correct because it includes the negative sign. Option B, 4, has the wrong sign: it would describe a sequence increasing by 4. Option C, 5, is only the constant part of the expression; it is not the change from one term to the next. Option D, \(-5\), confuses the constant 5 with the coefficient of \(n\). A negative common difference means the terms go downward, not that the sequence is invalid. For a quick exam method, compare \(a_{n+1}\) and \(a_n\): \(a_{n+1}-a_n=[5-4(n+1)]-[5-4n]=-4\).

Given \(a_n=5n^2+2\), \(a_4=5(4)^2+2=5\times16+2=82\) and \(a_1=5(1)^2+2=7\). Therefore, \(a_4+a_1=82+7=89\). Option 87 can result from evaluating \(4^2\) incorrectly while finding \(a_4\). Exam tip: substitute the value of \(n\) separately for each term before adding them.

The general term is (a_n=68-8n), and (68-8n=-4) gives (n=9). Even in decreasing sequences the position is natural.

Given \(a_n=3n^2-2n+5\), substitute \(n=1\), \(2\), and \(3\): \(a_1=3(1)^2-2(1)+5=6\), \(a_2=3(2)^2-2(2)+5=13\), and \(a_3=3(3)^2-2(3)+5=26\). Thus, the first three terms are \((6, 13, 26)\). Option D has 14 as the second term, but correct substitution gives 13. Exam tip: substitute each value of \(n\) separately, applying the square and multiplication first.

The numerator is (n^2+4) and the denominator is (n^2+1), so (a_n=\frac{n^2+4}{n^2+1}). In a fractional sequence identify square patterns separately.

An explicit rule gives the value of any term directly from its position number. Here the numerators are 2, 4, 6, and 8, so they are obtained by multiplying the term number by 2. The denominators are 5, 8, 11, and 14; they increase by 3 and can be written as 3n+2. Therefore the numerator is 2n and the denominator is 3n+2, giving the rule in option B.

To check it, substitute n=1: \(a_1=\frac{2(1)}{3(1)+2}=\frac{2}{5}\). For n=2, the result is \(\frac{4}{8}\); for n=3, it is \(\frac{6}{11}\); and for n=4, it is \(\frac{8}{14}\). All listed terms agree. Option A has the right denominator but misses the factor 2 in the numerator, while the other choices do not reproduce the terms.