Explicit or general rule

The first five terms are (3,16,29,42,55), and the average is (29). In exams, divide the sum by the number of terms.

Its rule is (a_n=13n-10), and (13n-10=159) gives (n=13). In exams, equate the given term to the general term.

Substitute n=3: a_3=3^4-3^2=81-9=72. Therefore, 72 is the correct option. A value such as 70 can result from an error while evaluating the powers or subtracting. Exam tip: calculate each power separately before subtracting.

For \(a_n=n^4-n^2\), we get \(a_1=1-1=0\), \(a_2=16-4=12\), \(a_3=81-9=72\), and \(a_4=256-16=240\). Hence, the correct general term is \(a_n=n^4-n^2\). The close distractor \(2n^4\) gives \(a_1=2\), which does not match the first term 0. Exam tip: substitute \(n=1,2,3\) to verify a proposed general term against the initial terms.

Given \(a_n=4^n+n-3\), substitute \(n=4\): \(a_4=4^4+4-3=256+4-3=257\). Therefore, 257 is correct. A value such as 255 can result from an error while evaluating the \(n-3\) part. Exam tip: after substitution in a general term, evaluate the exponent first and then perform addition and subtraction carefully.

Substituting \(n=1,2,3,4\) into \(a_n=4^n+n-3\) gives \(2,15,64,257\), respectively. Although \(a_n=n^4+1\) gives the first term as 2, its second term is \(17\), not 15. Exam tip: verify an explicit rule by testing it for at least the first three terms.

Given \(a_n=2n^2+9n+1\). Substituting \(n=5\), \(a_5=2(5)^2+9(5)+1=2\times25+45+1=96\). Hence, \(96\) is the correct option. A value such as \(94\) can result from an error while evaluating the squared term \(2\times25\). Exam tip: substitute the value of \(n\) first, then evaluate the exponent carefully.

The given rule is \(a_n=200-12n\). Substituting \(n=9\), we get \(a_9=200-12\times9=200-108=92\). Therefore, 92 is correct. A value such as 88 results from an arithmetic error after substitution. Exam tip: substitute the term number first, then perform multiplication before subtraction.

At (n=1) it gives (188), and at (n=2) it gives (176), so (a_n=200-12n). In exams, check the first two terms of a decreasing sequence.

Substitute \(n=6\): \(a_6=\frac{6(5\times6-1)}{2}=\frac{6(30-1)}{2}=\frac{6\times29}{2}=87\). Hence, the correct answer is 87. A value such as 84 can result from an error while evaluating \(5n-1\). Exam tip: substitute the term number first, then simplify the bracket and perform multiplication and division step by step.

Given \(a_n=(n+1)^3-n\), substitute \(n=4\): \(a_4=(4+1)^3-4=5^3-4=125-4=121\). Therefore, 121 is the correct option. 123 would result from an incorrect subtraction. Exam tip: substitute the term number first, evaluate the exponent, and then subtract.

Substituting \(n=1,2,3,4\) into \(a_n=(n+1)^3-n\) gives \(7,25,61,121\), respectively. Therefore, option A is correct. Option B gives the first term as \(7\), but for \(n=2\) it gives \(14\), not \(25\). Exam tip: verify a proposed general term using at least the first three terms.

Substituting \(n=4\), \(a_4=3\cdot2^4+2(4)^2=3\cdot16+2\cdot16=48+32=80\). Therefore, 80 is correct. An answer such as 76 can result from incorrectly evaluating either the exponential or squared term. Exam tip: calculate \(2^4\) and \(4^2\) separately before multiplying by their coefficients.

The correct rule is \(a_n=3\cdot2^n+2n^2\). Checking it: for \(n=1\), \(3\cdot2+2=8\); for \(n=2\), \(3\cdot4+8=20\); and for \(n=3\), \(3\cdot8+18=42\). Option C gives the first term \(8\), but for \(n=2\) it gives \(16\), not \(20\). Exam tip: test an explicit rule using at least the first three terms.

Given \(a_n=7n^2-4n+1\), substitute \(n=5\): \(a_5=7(5)^2-4(5)+1=7\times25-20+1=175-20+1=156\). Hence, 156 is correct. The value 154 may result from incorrectly omitting the final \(+1\). Exam tip: substitute the term number first, then evaluate powers and multiplication carefully.

The direct answer is option A: \(a_n=7n^2-4n+1\). Check each position carefully. At \(n=1\), \(7-4+1=4\). At \(n=2\), \(7(4)-8+1=21\). At \(n=3\), \(63-12+1=52\). At \(n=4\), \(112-16+1=97\). Thus the formula gives all the listed terms. The first differences are 17, 31 and 45; the second differences are 14 and 14, indicating a quadratic expression. Option B, \(4n^2\), gives 4, 16, 36, 64, so only the first term is right. Option C, \(17n-13\), gives 4, 21, 38, 55; it assumes a constant difference, which is not present. Option D gives 7, 24, 51, 88, so it does not fit even the first term. Exam cue: calculate first and second differences, then verify the proposed formula by substitution.

The consecutive differences are \(19,29,39\), whose second differences are \(10\). Therefore, the sequence has a quadratic general term of the form \(a_n=5n^2+bn+c\), since its second difference is \(2\times5=10\). Substituting \(n=1\) and \(n=2\) gives \(b=4\) and \(c=-1\), so \(a_n=5n^2+4n-1\) is correct. The close distractor \(a_n=5n^2+2n+1\) gives the first term correctly but gives \(25\), not \(27\), for the second term. Exam tip: verify a proposed general term using at least the first two or three terms.

Using \(a_n=\frac{n(3n+5)}{2}\), substitute \(n=8\): \(a_8=\frac{8(3\times8+5)}{2}=\frac{8(29)}{2}=4\times29=116\). Hence, option C is correct. A value such as \(112\) can result from an error while evaluating the expression inside the bracket. Exam tip: substitute the value of \(n\) first, then simplify brackets and multiplication/division step by step.

The direct answer is option B: \(a_n=2\cdot5^{n-1}+n^2\). Substitute the position one by one. For \(n=1\), \(2\cdot5^0+1^2=2+1=3\). For \(n=2\), \(2\cdot5^1+2^2=10+4=14\). For \(n=3\), \(2\cdot5^2+3^2=50+9=59\). For \(n=4\), \(2\cdot5^3+4^2=250+16=266\). Every term matches. Option A gives 3, 12, 27, 48, so it fails at the second term. Option B correctly combines exponential growth with the added square term. Option C, \(5^n-n\), gives 4, 23, 122, 621, not the sequence. Option D, \(n^3+2n\), gives 3, 12, 33, 72, so only its first term matches. The important method is not to guess from rapid growth; substitute several values of \(n\). Memory cue: \(5^{n-1}\) starts with 1 at \(n=1\), while the separate \(n^2\) term must also be included.

From the given terms, (p+q=5) and (2p+q=7), so (p=2), (q=3), and (a_4=45). When coefficients are unknown, first form equations using small terms.