Explicit or general 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.

This is (5^3,6^3,7^3,8^3,\ldots), so (a_n=(n+4)^3). In cube sequences observe the order of base numbers.

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.

In a linear rule the coefficient of (n) is (-4), so the common difference is (-4). Do not miss the sign while reading the coefficient.

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.