Explicit or general rule

Use the given general rule \(a_n=11n-6\) separately for the two required positions. For \(n=7\), \(a_7=11(7)-6=77-6=71\). For \(n=2\), \(a_2=11(2)-6=22-6=16\). The requested expression is \(a_7-a_2\), so its value is \(71-16=55\). This also agrees with the fact that moving from the second to the seventh position covers five steps of size 11.

The answer must be the difference, not the value of \(a_7\) alone or the value of \(a_2\) alone. Careful substitution and subtraction lead to 55. Since 55 appears in option B, option B is correct. The other numerical choices do not result from the stated rule and the requested subtraction.

The magnitude is (5n) and signs start positive and alternate, so (a_n=(-1)^{n+1}5n). Choose the power of ((-1)) by checking the first term sign.

(3n^2+4n) does not give the sequence; the correct rule is (4n^2+3). Options should be matched with the first four terms.

Use the explicit rule at n = 4 because a₄ is given. Substitution gives a₄ = 4² + 4d + 2 = 16 + 4d + 2 = 18 + 4d. Setting this equal to 34 gives 18 + 4d = 34, so 4d = 16 and d = 4. Therefore, option C is correct. The other options result from omitting the constant 2 or making an error while solving the linear equation.

For a term to be negative, \(15-3n<0\). This gives \(n>5\), so the smallest integer value is \(n=6\). In fact, \(a_5=15-3(5)=0\), which is not negative, whereas \(a_6=15-3(6)=-3\). Hence, the sixth term is the first negative term. Exam tip: when asked for the first such term, choose the smallest natural-number index satisfying the inequality.

The increase over five gaps is (25), so (d=5), hence (a_{12}=42+3(5)=57). Extend the terms using the common difference.

For option D, substituting \(n=1,2,3,4\) gives \(10,21,34,49\), respectively. Hence, the correct general term is \(a_n=n^2+8n+1\). Option A gives the first term as 10, but for \(n=2\) it gives 20, not 21. Exam tip: When first differences are not constant, use the second differences to test a quadratic form \(an^2+bn+c\).

To find the fifth term, substitute \(n=5\) in the rule: \(a_5=(2\times5-1)^2+2=9^2+2=81+2=83\). Therefore, 83 is the correct option. The value 81 is only \(9^2\); the additional 2 in the rule must also be included. Exam tip: substitute the value of \(n\) first, then simplify brackets, powers, and addition in order.

The change over five gaps is (-15), so (d=-3), hence (a_n=21-3n). From two given terms first find the common difference.

The first differences are \(7,11,15\), and the second difference is constant at \(4\), so the general term should be quadratic. Substituting \(n=1,2,3,4\) in \(a_n=2n^2+n\) gives \(3,10,21,36\), respectively. Option A gives the first term as \(3\), but its second term is \(8\), not \(10\). Exam tip: verify a proposed general term using at least the first three terms.

In option A, the first differences are \(3,5,7,9\), so the second difference is constantly \(2\). A constant second difference indicates a quadratic rule. Option B has constant first differences and is linear. Exam tip: make a difference table to classify a sequence.

The correct rule is \(a_n=3n^2+2n\). Checking it gives \(a_1=3+2=5\), \(a_2=12+4=16\), and \(a_3=27+6=33\). The linear rule \(11n-6\) may match the first two terms, but for \(n=3\) it gives 27, not 33. In an exam, test an explicit rule with at least three terms.

(a_5=98) and (a_2=11), so the difference is (87). In exams, find both terms correctly before subtracting.

The first differences are \(7,13,19\), and the second differences are constant: \(6,6\). Hence, the general term should be quadratic. Substituting \(n=1,2,3,4\) in \(a_n=3n^2-2n+1\) gives \(2,9,22,41\), respectively. Option C matches the first two terms but gives \(18\), not \(22\), when \(n=3\). Exam tip: constant second differences usually indicate a rule of the form \(an^2+bn+c\).

(a_7=\frac{7\times10}{2}+2=37). In exams, simplify the fraction first and then add the constant.

The terms \(4,9,16,25\) are respectively \(2^2,3^2,4^2,5^2\). When \(n=1\), the number being squared is \(2=n+1\); hence the general term is \(a_n=(n+1)^2\). Although \(a_n=n^2+3\) gives 4 for the first term, it gives 7, not 9, when \(n=2\). Exam tip: verify a general term by substituting \(n=1\) and \(n=2\).

Given \(a_n=(2n-1)^2\). Substituting \(n=5\), we get \(a_5=(2\times5-1)^2=(10-1)^2=9^2=81\). Hence, 81 is the correct option. The value 64 would require the expression inside the square to be 8, which it is not. Exam tip: substitute the value of \(n\) first, and then evaluate the exponent.

The governing concept is evaluating multiple terms from an explicit quadratic sequence rule and then combining them. First calculate a₄ by substituting n = 4: a₄ = 2(4²) + 5(4) − 3 = 2(16) + 20 − 3 = 49. Next calculate a₂: a₂ = 2(2²) + 5(2) − 3 = 2(4) + 10 − 3 = 15, not 5. Therefore a₄ + a₂ = 49 + 15 = 64. Since 64 is not among the supplied options, the item is internally inconsistent and cannot have a valid unambiguous answer as written. The stated key B and its explanation are mathematically incorrect; option B would require a₂ = 5, which does not follow from the rule.

(4n^2-n+1) gives (4,15,34,61). In exams, check a quadratic rule when second differences are constant.

\(a_{n+1}-a_n=[12-4(n+1)]-(12-4n)=-4\), which is constant for every \(n\). Hence it is an arithmetic progression with common difference \(-4\), not \(4\). Exam tip: check the difference of consecutive terms.