Explicit or general rule

Its rule is (a_n=3n^2+4n+4), so the (10)th term is (344), not any listed option. In exams also check the consistency of options.

Substituting \(n=1,2,3,4\) into \(a_n=n^2+6n+1\) gives \(8,17,28,41\), respectively. Therefore, option A is correct. In option B, the term for \(n=2\) is \(18\), not the given second term \(17\). Exam tip: Verify a proposed sequence rule by substituting at least the first two or three values of \(n\).

The general term is (a_n=8n+5), and (8n+5=93) gives (n=11). If (n) is a natural number, the given term belongs to the sequence.

Substitute \(n=3\) in the given rule: \(a_3=6^3-3(3)=216-9=207\). Therefore, 207 is the correct option. The value 213 would result from not subtracting the \(3n\) term, so it is incorrect. Exam tip: For a term of a sequence, substitute the value of \(n\) everywhere first, then evaluate powers and the remaining operations.

The numerator is (3n+2) and the denominator is (4n+2), so (a_n=\frac{3n+2}{4n+2}). In a fractional sequence form separate rules for both parts.

The increase over five gaps is (35), so (d=7), then (a_n=7n-5). From two given terms first find the common difference.

Substituting n=1, 2, and 3 into the rule gives a_1=3(1)^2-2(1)+6=7, a_2=3(2)^2-2(2)+6=14, and a_3=3(3)^2-2(3)+6=27. Hence, the correct sequence is (7, 14, 27). Option D has the correct first term, but for n=2 the value is 14, not 16. Exam tip: substitute each value of n carefully into the entire expression.

This is three times the triangular numbers, so (a_n=\frac{3n(n+1)}{2}). Recognize the pattern from additions (6,9,12).

For option A, substituting \(n=1,2,3,4\) gives \(12,25,42,63\), respectively. Also, the first differences are \(13,17,21\), so the second differences are constant at \(4\); this supports a quadratic rule. Option C matches the first two terms but gives \(40\), not \(42\), when \(n=3\). Exam tip: test a proposed rule with at least three terms, not just the first term.

From (\frac{3n}{n+4}=\frac{9}{7}), (21n=9n+36), so (n=3). None of the given options is correct.

Given \(a_n=4n^2+3\), \(a_3=4(3)^2+3=39\) and \(a_5=4(5)^2+3=103\). Therefore, \(a_3+a_5=39+103=142\). A value such as 140 can result from an error while squaring or adding. Exam tip: substitute each required value of \(n\) separately into the general term before finding their sum.

This is (4^2,8^2,12^2,16^2,\ldots), so (a_n=(4n)^2=16n^2). In square sequences with equal base gaps, find the base rule.

The magnitude is (6,11,16,21,\ldots) and signs start positive and alternate, so (a_n=(-1)^{n+1}(5n+1)). The sign of the first term decides the power.

The change over six gaps is (-36), so (d=-6), hence (a_n=31-6n). From two given terms first find the common difference.

Given \(n^2+5n=126\), we get \(n^2+5n-126=0\). Factoring gives \((n-9)(n+14)=0\), so \(n=9\) or \(n=-14\). Since a term position \(n\) in a sequence is positive, \(n=9\) is correct. For example, \(n=8\) gives 104, not 126. Exam tip: bring all terms to one side and factor the quadratic expression.

The first differences are \(4,6,8\), and the second differences are constant: \(2,2\). Hence, the general term should be quadratic. Substituting \(n=1,2,3,4\) in \(a_n=n^2+n+2\) gives \(4,8,14,22\), respectively. Although \(n^2+3\) gives the first term as \(4\), it gives \(7\) as the second term, so it is incorrect. Exam tip: when second differences are constant, test a quadratic rule using the first few terms.

The direct answer is option C: \(\frac{17}{11}\). To find \(a_4\), replace every \(n\) in the numerator and denominator by 4. The numerator is \(4(4)+1=16+1=17\). The denominator is \(3(4)-1=12-1=11\). Therefore \(a_4=\frac{17}{11}\). The denominator is not zero, so the value is valid. Option A, \(\frac{15}{11}\), would use the wrong numerator because the numerator is 17, not 15. Option B, \(\frac{16}{11}\), leaves out the added 1 in the numerator. Option C has both substitutions correct. Option D, \(\frac{18}{11}\), adds too much in the numerator and does not follow \(4n+1\). The essential habit is to substitute the same index into both parts of a fraction. Memory cue: write the numerator and denominator on separate lines before simplifying, so signs and constants are not missed.

The direct answer is option B: \(a_n=(n+5)^3\). Recognise the terms as cubes: \(216=6^3\), \(343=7^3\), \(512=8^3\), and \(729=9^3\). The bases are 6, 7, 8, 9, which increase by 1. For the first term, \(n=1\), so the base must be \(1+5=6\); therefore the general base is \(n+5\), and cubing it gives the rule. Option A, \((n+4)^3\), starts with \(5^3=125\), so it is wrong. Option B gives \(6^3,7^3,8^3,9^3\), exactly the sequence. Option C gives \(n^3+215\): its first terms are 216, 223 and 242, so only the first matches. Option D gives 6, 48, 162 and 384, not the given terms. Memory cue: in a cube sequence, first identify the cube roots, then match their position with \(n\).

In option A, substituting \(n=1,2,3,4\) gives \(14,31,54,83\), respectively. Therefore, the correct rule is \(a_n=3n^2+8n+3\). For example, when \(n=3\), \(a_3=3(3)^2+8(3)+3=54\). Option B also gives 14 as its first term, but it gives 29, not 31, when \(n=2\). Exam tip: verify at least the first two or three terms before selecting a general rule.

The direct answer is option A: \(-5\). The sequence rule is \(a_n=11-5n\). In a linear general term, the coefficient of \(n\) gives the change when the term number increases by 1. We can verify this carefully. The first term is \(a_1=11-5=6\), and the second term is \(a_2=11-10=1\). Their difference is \(1-6=-5\). The third term is \(a_3=11-15=-4\), and \(-4-1=-5\) again. Thus the sequence decreases by 5 at every step. Option A is correct. Option B, 5, has the opposite sign and would describe increasing terms. Option C, 11, is the constant term in the formula, not the common difference. Option D, \(-11\), incorrectly treats the constant 11 as the changing part. Another algebraic check is \(a_{n+1}-a_n=[11-5(n+1)]-[11-5n]=-5\). A negative difference is perfectly valid; it simply means the terms move downward. The common mistake is to read the first number in the formula instead of the coefficient attached to \(n\).

The general term is (a_n=79-9n), and (79-9n=-2) gives (n=9). Even in decreasing sequences the position is natural.

Substitute n=1, 2, and 3 successively. This gives a_1=4(1)^2-3(1)+8=9, a_2=4(2)^2-3(2)+8=18, and a_3=4(3)^2-3(3)+8=35. Hence, the correct sequence is (9,18,35). Option (9,20,39) has the correct first term, but its calculations for n=2 and n=3 are incorrect. Exam tip: substitute each value of n separately, carrying out squaring and multiplication first.

The numerator is (n^2+5) and the denominator is (n^2+4), so (a_n=\frac{n^2+5}{n^2+4}). Identify square patterns in fractional sequences.

The rule of a sequence can be found by observing how the numerator and denominator depend on n. The numerators are 3, 6, 9, and 12, which are exactly 3n for n=1, 2, 3, and 4. The denominators are 7, 12, 17, and 22. They increase by 5, and the expression 5n+2 gives these values. Thus the general term is \(a_n=\frac{3n}{5n+2}\), which is option A.

Substitution verifies the answer: at n=1, \(\frac{3}{7}\) is obtained; at n=2, \(\frac{6}{12}\); at n=3, \(\frac{9}{17}\); and at n=4, \(\frac{12}{22}\). Option C uses the wrong denominator pattern, and options B and D do not produce the numerators 3n. Both parts of the fraction must match the sequence.

The magnitude is (7n) and signs start positive and alternate, so (a_n=(-1)^{n+1}7n). Choose the power of ((-1)) by checking the first term sign.