Explicit or general rule

(a_5=2) and (a_6=-2), so the first negative term is the (6)th. Do not count zero or positive terms as negative.

The increase over six gaps is (54), so (d=9), hence (a_{15}=83+4(9)=119). Extend the terms using the common difference.

With only four given terms, the general term of a sequence is not uniquely determined. For example, the polynomial in option C gives \(1,15,63,195\) for \(n=1,2,3,4\), but many other rules can also produce these same four terms. Option A fits only the first two terms; at \(n=3\), it gives \(53\), not \(63\). Exam tip: Test a proposed rule against every listed term and check whether enough information is available to define a unique pattern.

When \(n\) increases by 1 in \(a_n=4n-1\), the term increases by 4, so \(a_{n+1}-a_n=4\) is constant. Hence it is an arithmetic progression, not a GP, which needs a constant ratio. Exam tip: in \(pn+q\), the common difference is \(p\).

For \(a_n=5n^2+2n\), substituting \(n=1,2,3,4\) gives \(7,24,51,88\), respectively. Hence, it is the required explicit rule. Although \(a_n=3n^2+4n\) gives the first term as 7, it gives 20, not 24, when \(n=2\). Exam tip: test a rule using at least the first three terms.

The consecutive differences are 5−2=3, 8−5=3, and 11−8=3, so the common difference is constant and the sequence is an AP. In option B, the differences 1, 2, 4 change. Exam tip: check consecutive differences to identify an AP.

The successive differences are \(9,15,21\), and their second differences are constant: \(6,6\). Hence the rule is quadratic, with coefficient of \(n^2\) equal to \(6/2=3\). Substituting in \(a_n=3n^2-1\) gives \(a_1=2, a_2=11, a_3=26\), and \(a_4=47\). Option D is linear and would require equal first differences. Exam tip: constant second differences usually indicate a quadratic sequence rule.

Substitute \(n=5\): \(a_5=\frac{5(4\times5+1)}{2}=\frac{5(21)}{2}=\frac{105}{2}\). Hence, option C is correct. \(\frac{101}{2}\) may result from evaluating \(4\times5+1\) incorrectly. Exam tip: after substituting the term number in a general term, follow the order of operations carefully.

Given \(a_n=(3n-2)^2\). Substituting \(n=4\), \(a_4=(3\times4-2)^2=(12-2)^2=10^2=100\). Therefore, 100 is the correct option. \(121\) would result only if the expression inside the square were 11, which it is not here. Exam tip: substitute the value of \(n\) first, then simplify the bracket before applying the exponent.

The terms are \(1^2,4^2,7^2,10^2\). Their square roots, \(1,4,7,10\), form an arithmetic sequence whose \(n\)th term is \(3n-2\). Hence the explicit rule is \(a_n=(3n-2)^2\). In option B, substituting \(n=2\) gives 10, not 16. Exam tip: for sequences of perfect squares, first check the pattern in their square roots.

Given \(a_n=150-11n\), substitute \(n=7\): \(a_7=150-11\times7=150-77=73\). Hence, \(73\) is correct. \(77\) is only the value of \(11\times7\), not the value of \(a_7\). Exam tip: substitute the term number carefully, multiply first, and then subtract.

At (n=1) it gives (139), and at (n=2) it gives (128), so (a_n=150-11n). In exams, match the first term of a decreasing sequence.

Its rule is (a_n=11n-7), and (11n-7=125) gives (n=12). In exams, equate the given term to the general term.

The term number starts with \(n=1\). Substituting \(n=1,2,3,4\) into \(a_n=2^n+n^2\) gives \(3,8,17,32\), respectively. The close distractor \(2n^2+1\) gives \(3,9\) as its first two terms, so it does not fit the sequence. Exam tip: verify a proposed general term using at least the first three values of \(n\).

Putting n=3, we get a_3=3^3+3^2-1=27+9-1=35. Therefore, the correct answer is 35. A value such as 33 results from an incorrect calculation of the final subtraction; 27+9-1 equals 35. Exam tip: when finding a term from an explicit rule, substitute the given value of n in every occurrence of n, including powers and squares.

The correct rule is \(a_n=3^n+n^2-1\). Substituting \(n=1,2,3,4\) gives \(3,12,35,96\), respectively. Although \(a_n=3n^2\) gives the first two terms as 3 and 12, it gives 27 as the third term instead of 35, so it is not the rule for the sequence. In exams, test a proposed general term with at least the first three terms.

Substitute \(n=5\): \(a_5=7\cdot2^{5-1}+4=7\cdot2^4+4=7\cdot16+4=116\). Therefore, the correct answer is 116. The value 112 is only \(7\cdot16\); it misses the final \(+4\). Exam tip: evaluate \(n-1\) first, then calculate the power.

Given \(a_n=4n^2+5n-6\), \(a_3=4(3)^2+5(3)-6=45\) and \(a_5=4(5)^2+5(5)-6=119\). Therefore, \(a_3+a_5=45+119=164\). A nearby value such as 162 can result from a small addition or multiplication error. Exam tip: substitute each value of \(n\) separately before adding the terms.

The direct answer is option A: \(a_n=4n^2+5n-6\). Start by checking the terms. For \(n=1\), this gives \(4+5-6=3\); for \(n=2\), it gives \(16+10-6=20\); for \(n=3\), it gives \(36+15-6=45\); and for \(n=4\), it gives \(64+20-6=78\). Thus it matches every supplied term. The first differences are \(17,25,33\), and the second differences are \(8,8\), so a quadratic rule is reasonable. Option A is exactly the matching rule. Option B, \(3n^2\), gives 3, 12, 27, 48, so it fails after the first term. Option C, \(17n-14\), is linear and gives 3, 20, 37, 54, so it fails at the third term. Option D, \(5n^2-2n\), gives 3, 16, 39, 72, so it also fails. A useful check is always to substitute \(n=1,2,3\), not merely trust the appearance of a formula. Memory cue: constant second differences usually indicate a quadratic sequence.

Substitute n=4: \(a_4=4^3+3(4^2)-2=64+48-2=110\). Therefore, 110 is the correct option. A nearby distractor such as 108 results from an incorrect final calculation; \(3\times4^2\) must be evaluated as 48. Exam tip: calculate powers first, then perform multiplication and addition or subtraction.

For \(a_n=n^3+3n^2-2\), substituting \(n=1,2,3,4\) gives \(2,18,52,110\), respectively. Hence, it is the correct general term. The close distractor \(2n^3\) gives the first term as 2, but gives 16 rather than 18 when \(n=2\). In an exam, verify a proposed general term using at least the first two or three terms.

Substitute \(n=3\) in the given rule: \(a_3=2(3)^3+(3)^2+3=2\times27+9+3=54+9+3=66\). Therefore, 66 is the correct option. Getting 64 indicates an error in evaluating the cubic term \(2(3)^3\). Exam tip: substitute the value in every term first, then evaluate powers before adding.

The correct rule is \(a_n=2n^3+n^2+n\). Substituting \(n=1,2,3,4\) gives \(4,22,66,148\), respectively. The closest distractor, \(4n^2\), gives 4 when \(n=1\), but it gives 16 rather than 22 when \(n=2\). Exam tip: verify a proposed general term using at least the first three terms.

Substitute \(n=4\): \(a_4=\frac{3(4)^2+5(4)}{4}=\frac{3\times16+20}{4}=\frac{68}{4}=17\). Therefore, 17 is correct. Getting 16 may result from incorrectly evaluating \(3\times4^2\) or making an error in the numerator. Exam tip: substitute the value of \(n\) first, evaluate the power, and then simplify the numerator and denominator.

Substitute 3 for n in the general term: \(a_3=5^3+3-4=125+3-4=124\). Therefore, the correct answer is 124. A value such as 122 can result from an arithmetic error in the addition or subtraction. Exam tip: evaluate the power first, then add or subtract the remaining terms.