Explicit or general rule

The governing concept is solving an explicit term equation for the position n. Set the formula equal to the target value: n^2 + 6n + 5 = 252. Rearranging gives n^2 + 6n − 247 = 0. The factors of −247 that have sum 6 are 19 and −13, so the quadratic factors as (n + 19)(n − 13) = 0. Thus n = −19 or n = 13. A term position must be positive, so n = 13 is valid. Direct verification gives a_13 = 13^2 + 6(13) + 5 = 169 + 78 + 5 = 252. Therefore option C is correct. The negative root is rejected because sequence positions are positive integers; the other options do not satisfy the formula.

The direct answer is option A, \(a_n=5n+2\). This is an arithmetic sequence because the difference is constant: \(12-7=5\), \(17-12=5\), and \(22-17=5\). For an arithmetic sequence, use \(a_n=a_1+(n-1)d\). Here \(a_1=7\) and \(d=5\), so \(a_n=7+(n-1)5=7+5n-5=5n+2\). Checking: for \(n=1\), it gives 7; for \(n=2\), 12; for \(n=3\), 17; and for \(n=4\), 22. Option A works exactly. Option B gives 12 as its first term, so it starts at the second given term. Option C gives 12, 19, 26, so its differences and starting value are wrong. Option D gives 7, 19, 31, so it does not match the sequence. Memory cue: first term plus repeated difference means \(a_1+(n-1)d\), not \(a_1+nd\).

For \(1,4,9,16,\ldots\), \(a_n=n^2\), so the rule has degree 2. The first differences are 3, 5, 7, giving a constant second difference of 2. Exam tip: constant second differences indicate a quadratic sequence.

For (n=1,2,3,4), (2n^2+1) gives the given terms. While checking options, match the first four terms.

((-1)^n) gives (-1) for odd (n) and (1) for even (n). In exams, check odd-even positions in alternating rules.

This is an arithmetic progression because the difference between consecutive terms is 5. Here, \(a=-2\), \(d=5\), and \(n=20\). Therefore, \(a_{20}=a+(n-1)d=-2+(20-1)\times5=-2+95=93\). Hence, 93 is correct. Getting 98 usually results from miscounting the terms or using an incorrect value in place of \(n-1\). Exam tip: for an arithmetic progression, always use \(a_n=a+(n-1)d\).

The common difference is \(d=\frac{a_8-a_3}{8-3}=\frac{39-14}{5}=5\). Hence, \(a_n=a_3+(n-3)d=14+5(n-3)=5n-1\). Therefore, \(5n-1\) is correct. Although \(4n+2\) gives \(a_3=14\), it gives \(a_8=34\), not 39. Exam tip: First find \(d=\frac{a_q-a_p}{q-p}\) from two known terms, then substitute either known term.

The terms are \(1^2, 2^2, 3^2, 4^2,\ldots\). Therefore, the \(n\)th term is \(a_n=n^2\). If \(a_n=n^2+1\), the first term would be 2, which does not match the sequence. Exam tip: write the first few terms alongside \(n=1,2,3,\ldots\) to check the pattern.

Given \(a_n=2^n+1\), substitute \(n=6\): \(a_6=2^6+1=64+1=65\). Hence, 65 is the correct option. The value 63 would result from \(2^n-1\), so it is not correct here. Exam tip: In a general term of a sequence, substitute the given value of \(n\) first and then evaluate the exponent.

This is an arithmetic progression with first term \(a=10\) and common difference \(d=7-10=-3\). Therefore, the \(n\)th term is \(a_n=a+(n-1)d=10+(n-1)(-3)=13-3n\). In \(10-3n\), putting \(n=1\) gives 7 rather than the first term 10, so it is incorrect. Exam tip: verify a general term by substituting \(n=1\); it must give the first term.

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).

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.

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.

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.