Explicit or general rule

Substituting \(n=1,2,3,4\) in \(n(n+1)\) gives \(1\cdot2=2\), \(2\cdot3=6\), \(3\cdot4=12\), and \(4\cdot5=20\). Hence, the general term is \(a_n=n(n+1)\). The close distractor \(n^2+1\) gives 5 as its second term, so it does not fit the sequence. Exam tip: test a proposed general rule by substituting \(n=1,2,3\) and matching the initial terms.

Given \(a_n=3n^2-2\), \(a_4=3(4)^2-2=46\) and \(a_5=3(5)^2-2=73\). Therefore, \(a_4+a_5=46+73=119\), so option C is correct. A value such as 115 can result from an error while squaring or multiplying for \(a_5\). Exam tip: substitute the value of \(n\), evaluate the square first, and then perform the remaining operations.

For option B, \(a_n=3n^2+1\) gives \(a_1=3(1)^2+1=4\), \(a_2=3(2)^2+1=13\), \(a_3=28\), and \(a_4=49\). Hence, it is the correct general term. For example, option A gives \(a_2=11\), which does not match the second term, 13. Exam tip: test a proposed rule for at least the first two or three terms.

To find the position whose value is 41, put a_n=41: 2n+5=41. Thus, 2n=36 and n=18. Therefore, 41 is the 18th term of the sequence. The 17th term is 2(17)+5=39, so it is not correct. Exam tip: To find a term’s position from an explicit rule, substitute the given term value for a_n and solve for n.

These are triangular numbers and the rule is (a_n=\frac{n(n+1)}{2}). Recognize the pattern of adding (1,2,3,4).

For the first negative term, we need \(10-2n<0\). This gives \(n>5\), so the smallest integer value is \(n=6\). Checking nearby terms: \(a_5=10-2(5)=0\), which is not negative, whereas \(a_6=10-2(6)=-2\). Exam tip: a negative term must be less than \(0\); zero is neither positive nor negative.

This option set has a mismatch because (n^2+5) gives (6,9,14,21), not the given sequence. The correct rule for (6,11,18,27) is (n^2+2n+3).

The governing concept is the general term of an arithmetic sequence. The consecutive differences are 13−7=6, 19−13=6, and 25−19=6, so the common difference is 6. For an arithmetic sequence, a_n=a_1+(n−1)d. Substituting a_1=7 and d=6 gives a_n=7+6(n−1)=7+6n−6=6n+1. Therefore option A is correct. Checking n=1 gives 7, n=2 gives 13, and n=4 gives 25, confirming the rule. Option C gives 5 for the first term, option B has an incorrect coefficient and fails at n=2, and option D increases by only 1 rather than by 6.

For \(a_n=5-2n\), the difference \(a_{n+1}-a_n=-2\) is constant, so it is arithmetic. In option B, the ratio of consecutive terms is constant, making it geometric. Exam tip: an AP rule is usually linear in \(n\), of the form \(pn+q\).

At (n=1) it gives (58) and at (n=2) it gives (51) so (a_n=65-7n). In exams treat the difference as negative in decreasing sequences.

The governing concept is an explicit or general rule: a formula gives the value of a sequence directly for any term number n. To find which term has value 75, set the formula equal to 75: 4n + 7 = 75. Subtracting 7 from both sides gives 4n = 68, and dividing by 4 gives n = 17. Therefore, the required term is the seventeenth term, so option C is correct. Options A, B, and D do not satisfy the equation; their substitution gives 67, 71, and 79 respectively, rather than 75.

Its rule is (a_n=8n+1) and (8n+1=81) gives (n=10). In exams recheck the calculation before choosing the option.

Substituting \(n=1,2,3,4\) in \(a_n=2n^2+n+2\) gives \(5,12,23,38\), respectively. Therefore, \(2n^2+n+2\) is the correct general term. The close distractor \(7n-2\) matches the first two terms but gives 19, not 23, for the third term. Exam tip: verify a proposed general term using at least the first three or four terms.

\(2n^2+3n\) is a quadratic rule, so its first differences are linear and its second differences are constant. For \(5n-1\), the first differences themselves are constant. Exam tip: constant second differences indicate a quadratic sequence.

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

This question tests the use of an explicit sequence rule. The expression aₙ = 5n + 3 gives the value of the nth term, so to find the term equal to 88 we solve 5n + 3 = 88. Subtract 3 from both sides to obtain 5n = 85. Dividing by 5 gives n = 17. Therefore, 88 is the seventeenth term and option C is correct. Checking confirms that a₁₇ = 5(17) + 3 = 85 + 3 = 88. The other choices produce 78, 83, and 93, respectively, so none of them meets the required value.

At (n=1) it gives (111) and at (n=2) it gives (102) so (a_n=120-9n). In exams keep the decreasing difference negative.

Given \(a_n=(n+2)^2-3\), substitute \(n=5\): \(a_5=(5+2)^2-3=7^2-3=49-3=46\). Therefore, 46 is the correct option. The nearby option 44 may result from an error while squaring or subtracting. Exam tip: substitute the value of \(n\) first, then evaluate brackets, powers, and the remaining operations in order.

For option A, substituting \(n=1,2,3,4\) gives \(6,13,22,33\), respectively. Hence, the correct general term is \(a_n=n^2+4n+1\). The closest distractor, option D, gives the first two terms as \(6,13\), but for \(n=3\) it gives \(20\), not \(22\). Exam tip: test a proposed general term using at least the first three values of \(n\).

Given \(a_n=\frac{n(n+4)}{2}\). For \(a_8\), substitute \(n=8\): \(a_8=\frac{8(8+4)}{2}=\frac{8\times12}{2}=48\). Hence, 48 is the correct option. Values such as 46 or 50 can result from an arithmetic error, but \(8+4=12\). Exam tip: Substitute the required value of \(n\) first, then simplify step by step.