Explicit or general rule

(a_3=72) and (a_{10}=30), so the sum is (102). In exams, find both terms carefully in a decreasing formula.

At (n=1) it gives (84), and at (n=2) it gives (78), so (a_n=90-6n). In exams, check the first two terms of a decreasing sequence.

Given \(a_n=\frac{n(2n+3)}{2}\). Substituting \(n=6\), \(a_6=\frac{6(2\times6+3)}{2}=\frac{6\times15}{2}=45\). Hence, 45 is correct. An answer such as 42 may result from evaluating \(2n+3\) incorrectly or making an error in multiplication or division. Exam tip: substitute the value of \(n\) first, then simplify the bracket and calculate step by step.

Substitute n=3: a_3=4^3-3^2=64-9=55. Therefore, 55 is the correct option. An answer such as 53 may result from an error in evaluating the power or the square. Exam tip: substitute the value of n in every part of the rule before calculating powers and squares.

The direct answer is option A: \(a_n=4^n-n^2\). Test the rule at each position. For \(n=1\), \(4^1-1^2=4-1=3\). For \(n=2\), \(4^2-2^2=16-4=12\). For \(n=3\), \(4^3-3^2=64-9=55\). For \(n=4\), \(4^4-4^2=256-16=240\). It reproduces every listed term. Option B, \(4n^2-1\), gives 3 first but 15 second, so it is a different quadratic rule. Option C, \(2^n+n\), gives 3 first but 6 second. Option D, \(n^4-1\), gives 0 first, not 3, and its growth is different. The correct reasoning is to test the proposed explicit formula at several positions, especially when powers and squares appear. Memory cue: calculate carefully; \(4^n\) means a power of 4, while \(4n^2\) means 4 multiplied by \(n^2\).

The first five terms are (5,17,29,41,53), and the average is (29). In exams, divide the sum by the number of terms.

Its rule is (a_n=12n-7), and (12n-7=149) gives (n=13). In exams, equate the given term to the general term.

Substituting \(n=2\) into the rule gives \(a_2=6(2)^2-4(2)+3=24-8+3=19\). Similarly, for \(n=3\), \(a_3=6(3)^2-4(3)+3=54-12+3=45\). Hence, option A is correct. Option B has an incorrect value of \(a_2\), while option C has an incorrect value of \(a_3\). Exam tip: substitute each value of \(n\) separately and simplify carefully.

Given \(a_n=5^n-2n\). Substituting \(n=3\), \(a_3=5^3-2(3)=125-6=119\). Therefore, 119 is correct. The value 121 may result from incorrectly subtracting 4 instead of \(2n=6\). Exam tip: substitute the required term number for \(n\) in every part of the rule.

Given \(a_n=2n^2+7n-4\), substitute \(n=6\): \(a_6=2(6)^2+7(6)-4=2\times36+42-4=72+42-4=110\). Therefore, the correct answer is 110. The value 108 may result from an error while calculating \(7\times6\). Exam tip: substitute the term number carefully and evaluate the power term first.

The correct general term is \(a_n=2n^2+7n-4\). Substituting \(n=1\), \(2\), and \(3\) gives \(a_1=5\), \(a_2=18\), and \(a_3=35\), respectively. The distractor \(13n-8\) may match the first two terms, but for \(n=3\) it gives \(31\), not \(35\). Exam tip: verify a proposed general term using at least the first three terms.

Substitute \(n=6\): \(a_6=\frac{2(6)^2+5(6)}{3}=\frac{2\times36+30}{3}=\frac{72+30}{3}=\frac{102}{3}=34\). Hence, 34 is the correct option. An answer such as 36 can result from not dividing the complete numerator correctly by 3. Exam tip: substitute the value of \(n\) first, then follow the order of operations—powers, multiplication, addition, and finally division.

The direct answer is option A: \(a_n=4n^2+3n+2\). Start with \(n=1\): \(4(1)^2+3(1)+2=9\). For \(n=2\), the value is \(4(2)^2+3(2)+2=24\); for \(n=3\), it is \(36+9+2=47\); and for \(n=4\), it is \(64+12+2=78\). Thus every displayed term is produced. The differences are 15, 23 and 31, and their second differences are both 8, which supports a quadratic rule. Option A works exactly. Option B, \(9n\), gives 9, 18, 27, 36, so it fails after the first term. Option C gives 9, 24, 49, 84, so its third and later terms are wrong. Option D gives 9, 24, 39, 54, a constant-difference sequence, not the given one. Memory cue: equal second differences suggest a quadratic expression; always substitute \(n=1,2,3,4\) to verify.

Substitute \(n=5\): \(a_5=3\cdot2^5-5=3\cdot32-5=96-5=91\). Therefore, the correct answer is 91. A value such as 89 can result from subtracting the wrong value instead of 5. Exam tip: evaluate the exponent first, then multiply, and finally subtract \(n\).

Substituting \(n=1,2,3,4\) into \(a_n=3\cdot2^n-n\) gives \(5,10,21,44\), respectively. Hence, option A is correct. Option C gives 3 when \(n=1\), so it does not even produce the first term. Exam tip: verify an explicit rule by checking at least the first three terms.

(a_5=99) and (a_2=8), so the sum is (107). In exams, handle signs carefully in a quadratic formula.

The first differences are \(14,22,30\), and the second differences are constant at \(8\). Hence the sequence is quadratic, with coefficient of \(n^2\) equal to \(8/2=4\). Substituting \(n=1,2,3,4\) in \(a_n=4n^2+2n\) gives \(6,20,42,72\), respectively. Option B matches the first term but gives \(16\), not \(20\), when \(n=2\). Exam tip: verify a proposed general term using at least the first three terms.

The first term is (5) and the common difference is (7), so (a_n=5+(n-1)7=7n-2). In an arithmetic sequence the coefficient is the common difference.

The first term is (42) and the difference is (-6), so (a_n=42+(n-1)(-6)=48-6n). Keep the difference negative for a decreasing sequence.

The rule is \(a_n=4n^2-5n+2\). Substituting \(n=6\), \(a_6=4(6)^2-5(6)+2=4\times36-30+2=116\). Therefore, 116 is the correct answer. A value such as 112 can result from an error while evaluating the squared term. Exam tip: calculate \(n^2\) first, then perform multiplication and the remaining operations.

Given \(a_n=9n+4\) and \(a_n=112\), set \(9n+4=112\). This gives \(9n=108\), so \(n=12\). Therefore, 112 is the 12th term of the sequence. The 11th term is \(9(11)+4=103\), so it is not correct. Exam tip: To find the position of a given term, equate the general term \(a_n\) to that value and solve for \(n\).

The numerator is (2n+1) and the denominator is (3n+2), so (a_n=\frac{2n+1}{3n+2}). In fractions observe numerator and denominator patterns separately.

For the fifth term, substitute \(n=5\): \(a_5=4\cdot3^{5-1}=4\cdot3^4=4\cdot81=324\). Therefore, 324 is correct. The value 216 can result from incorrectly using \(3^3\) and miscounting the exponent \(n-1\). Exam tip: after substituting the term number, evaluate \(n-1\) before calculating the power.

This is (5^2,6^2,7^2,8^2,\ldots), so (a_n=(n+4)^2). In square sequences relate the base number to (n).

(a_6=23) and (b_4=24), so the sum is (47). With two rules the term numbers may differ, so use them carefully.