Explicit or general rule

For \(a_n=3^n+2\), we get \(a_1=3+2=5\), \(a_2=9+2=11\), \(a_3=27+2=29\), and \(a_4=81+2=83\). Hence, the correct general term is \(a_n=3^n+2\). In contrast, \(a_n=3n+2\) grows linearly and gives \(11\) as its third term, not \(29\). Exam tip: test a proposed rule by substituting \(n=1,2,3\).

Given \(a_n=3^n+2\), substitute \(n=3\): \(a_3=3^3+2=27+2=29\). Hence, 29 is correct. The option 27 is only the value of \(3^3\); the added 2 has been omitted. Exam tip: substitute the term number first, then complete each operation in order.

Substitute \(n=5\) in \(a_n=5\cdot2^{n-1}\): \(a_5=5\cdot2^{5-1}=5\cdot2^4=5\cdot16=80\). Therefore, 80 is correct. Getting 40 may result from incorrectly using \(2^3\). Exam tip: first substitute the term number into the exponent \(n-1\), then calculate.

The first term is (5), and each term is multiplied by (2). In exams, check both the first term and ratio in a geometric sequence.

For \(a_n=5n-2\), \(a_{n+1}-a_n=[5(n+1)-2]-(5n-2)=5\), which is constant; hence it is an arithmetic progression. In \(n^2+1\), the differences vary. Exam tip: look for a linear rule \(pn+q\).

Substituting \(n=1,2,3,4\) in \(a_n=2n^2+3\) gives \(5,11,21,35\), respectively, so option C is correct. The close distractor \(a_n=3n^2+2\) gives the first term as 5, but for \(n=2\) it gives 14, not 11. Exam tip: test an explicit rule using at least the first three values of \(n\).

For \(a_n=5n-2\), \(a_{n+1}-a_n=5\) for every \(n\), so the common difference is constant. In \(n^2+1\), the differences change. Exam tip: test consecutive-term differences.

The first differences are \(8,14,20\), and the second differences are constant: \(6,6\). Hence, the general term should be quadratic. For \(a_n=3n^2-n\), substituting \(n=1,2,3,4\) gives \(2,10,24,44\), respectively. Option A gives \(8\) as the second term, while option D gives \(12\), so they are incorrect. Exam tip: test a proposed rule by substituting at least the first three values of \(n\).

Given \(a_n=n^3+n\), substitute \(n=3\): \(a_3=3^3+3=27+3=30\). Option 27 represents only \(3^3\); it misses the \(+n\) term in the rule. Exam tip: To find a term of a sequence, substitute the required index into every part of the formula.

Substituting \(n=1,2,3,4\) in \(a_n=n^3+n\) gives \(2,10,30,68\), respectively. Therefore, the correct general term is \(a_n=n^3+n\). The close distractor \(a_n=n^3+1\) gives the first term as 2, but for \(n=2\) it gives 9, not 10. Exam tip: verify a proposed general term using at least the first two or three terms.

Given \(a_n=2n-1\), put \(a_n=41\). Then \(2n-1=41\), so \(2n=42\) and hence \(n=21\). Therefore, 41 is the 21st term of the sequence. For example, if \(n=20\), then \(a_{20}=39\), not 41. Exam tip: To find the term number of a given value, substitute that value in the general term and solve for \(n\).

Its rule is (a_n=3n+4), and (3n+4=58) gives (n=18). In exams, write the rule first and compare it with the given term.

The difference is (6) and the first term is (8), so (a_n=6n+2). In exams, find the constant to match the first term.

Direct answer: A, 42. The rule is \\(a_n=6n+2\\), and the first three terms require n=1, 2 and 3. Step 1: \\(a_1=6(1)+2=8\\). Step 2: \\(a_2=6(2)+2=14\\). Step 3: \\(a_3=6(3)+2=20\\). Step 4: add them: \\(8+14+20=42\\). Therefore option A is correct. Option B, 40, is not the result of adding the three generated terms; it may come from an arithmetic mistake. Option C, 44, is also incorrect because the correct middle and final terms give a total of 42. Option D, 46, is incorrect for the same reason and does not follow from the formula. The important point is that the first three terms mean positions 1, 2 and 3, not the numbers 0, 1 and 2. Memory cue: for a sum of first terms, substitute each required position before adding.

Given \(a_n=4n-1\), \(a_7=4\times7-1=27\) and \(a_3=4\times3-1=11\). Therefore, \(a_7-a_3=27-11=16\). The option 14 would result from an incorrect subtraction of the terms. Exam tip: substitute the value of \(n\) and find each required term separately.

Given \(a_n=n^2+5n\), \(a_4=4^2+5(4)=16+20=36\) and \(a_2=2^2+5(2)=4+10=14\). Therefore, \(a_4:a_2=36:14=18:7\). Option (36:7) is not correct because only the first term has effectively been simplified. Exam tip: always reduce a ratio by dividing both terms by their greatest common factor.

For \(a_n=5n-2\), increasing \(n\) by 1 increases every term by 5, so the common difference is constant. In \(n^2+1\), differences vary. Exam tip: an AP rule has the linear form \(pn+q\).

Given \(a_n=n^2+n-1\), substitute \(n=6\): \(a_6=6^2+6-1=36+6-1=41\). Therefore, \(41\) is correct. The distractor \(40\) may result from an error in evaluating the expression or overlooking the final subtraction. Exam tip: after substituting a value in a general term, check the square, addition, and subtraction in order.

Given \(a_n=100-7n\), substitute \(n=8\): \(a_8=100-7\times8=100-56=44\). Therefore, option B is correct. Getting 42 would result from an incorrect calculation, since \(7\times8=56\). Exam tip: after substituting the term number, perform multiplication before subtraction.

At (n=1) it gives (93), and at (n=2) it gives (86), so (a_n=100-7n). In exams, match the first term of a decreasing sequence.

To find the fifth term, substitute \(n=5\): \(a_5=\frac{3(5)+1}{2}=\frac{16}{2}=8\). Hence, 8 is the correct option. A value such as 7 can result from an incorrect calculation of the numerator \(3n+1\). Exam tip: substitute the term number directly for \(n\) in the general-term formula.

(\frac{3n+1}{2}) gives (2,\frac{7}{2},5,\frac{13}{2}). In exams, match fractional terms using small (n) values.

Given \(a_n=\frac{n(n+1)}{2}\). Substituting \(n=6\), \(a_6=\frac{6(6+1)}{2}=\frac{6\times7}{2}=21\). Therefore, 21 is correct. A value such as 18 can result from using an incorrect value for \(n+1\). Exam tip: substitute the term number carefully and evaluate the brackets before simplifying.

Set the explicit formula equal to the target value: n(n + 1)/2 = 28. Multiplying both sides by 2 gives n(n + 1) = 56. We need two consecutive positive integers whose product is 56; they are 7 and 8. Hence n = 7, and direct substitution confirms it: a₇ = 7(7 + 1)/2 = 7×8/2 = 28. Therefore, option D is correct. Checking the distractors gives a₅ = 5×6/2 = 15, a₆ = 6×7/2 = 21, and a₈ = 8×9/2 = 36. These values show why the nearby indices 5, 6, and 8 do not work. The formula represents triangular numbers, and 28 is the seventh triangular number.

Substitute \(n=4\) in the general term: \(a_4=2^4+4=16+4=20\). Therefore, 20 is the correct option. A value such as 18 can result from an incorrect addition after evaluating \(2^4\). Exam tip: evaluate the exponent first, then perform the remaining operations.