Explicit or general rule

At (n=1) it gives (120), and at (n=2) it gives (112), so (a_n=128-8n). In exams, keep the decreasing difference negative.

The governing concept is using an explicit term rule in reverse to find the position of a specified value. Set the general expression equal to 146: 9n + 2 = 146. Subtract 2 from both sides to obtain 9n = 144, and divide by 9 to get n = 16. Hence the value 146 is the 16th term, so option C is correct. Verification gives a₁₆ = 9(16) + 2 = 144 + 2 = 146. Option B gives 137, while option D gives 155; option A gives 128. The essential step is solving the equation for the term number rather than substituting a guessed position into the sequence.

Its rule is (a_n=9n-3), and (9n-3=150) gives (n=17), so the correct term is the (17)th term.

Take \(n=1,2,3,4\). The values of \(n^2\) are \(1,4,9,16\); adding 1 gives \(2,5,10,17\). Hence, \(a_n=n^2+1\). The close distractor \(a_n=n^2+n\) gives 6 when \(n=2\), not 5. Exam tip: verify a proposed general term using at least the first three terms.

Given \(a_n=n^2+1\), set \(a_n=101\). Then \(n^2+1=101\), so \(n^2=100\) and \(n=10\). Hence, 101 is the 10th term of the sequence. Although the equation also gives \(n=-10\), a term number must be a positive integer, so it is rejected. Exam tip: move the constant term first, then take the square root of the resulting perfect square.

Given \(a_n=2^n+3n\), substitute \(n=5\): \(a_5=2^5+3(5)=32+15=47\). Option 43 may result from adding only 3 instead of evaluating \(3n\), which is incorrect. Exam tip: after substituting the value of \(n\), evaluate powers and multiplication before addition.

An explicit rule should produce every listed term when the position n is replaced by 1, 2, 3, and so on. The first term is obtained with n=1, the second with n=2, and the fourth with n=4. Testing several terms is important because a rule may match one value by coincidence. Here option A combines an exponential part \\(2^n\\) with a linear part \\(3n\\).

For option A, at n=1 we get \\(2^1+3(1)=5\\); at n=2, \\(2^2+3(2)=10\\); at n=3, \\(2^3+3(3)=17\\); and at n=4, \\(2^4+3(4)=28\\). These are exactly the given terms, so option A is correct. The other rules fail at one or more early positions and therefore cannot be the stated explicit rule.

Given \(a_n=3^n-2n+1\), substitute \(n=4\): \(a_4=3^4-2(4)+1=81-8+1=74\). Therefore, 74 is the correct answer. The value 72 would result from omitting the final \(+1\), so it is incorrect. Exam tip: After substituting the term number, evaluate the exponent and multiplication in brackets carefully.

Substituting \(n=1,2,3,4\) into \(a_n=3^n-2n+1\) gives \(2,6,22,74\), respectively. Hence, it is the general term. The close distractor \(2^n+n\) gives 3 as its first term, not 2. Exam tip: verify a proposed general term by checking it for at least the first three values of \(n\).

Substituting \(n=6\), \(a_6=5\cdot2^{6-1}-3=5\cdot2^5-3=5\cdot32-3=157\). Hence, 157 is correct. The close distractor 153 can result from mishandling the final \(-3\) or making an arithmetic error. Exam tip: substitute the term number into the exponent \(n-1\) first, then simplify step by step.

(5\cdot2^{n-1}-3) gives (2,7,17,37). In exams, also check constant subtraction in geometric forms.

Using a_n=3n^2+n-2: a_1=3(1)^2+1-2=2, a_2=3(2)^2+2-2=12, and a_3=3(3)^2+3-2=28. Therefore, the sum of the first three terms is 2+12+28=42. Option 40 is incorrect because the third term is 28, not 26. Exam tip: substitute each value of n separately before adding the terms.

The governing concept is selecting a quadratic general rule by using finite differences. The first differences are 12−2 = 10, 26−12 = 14, and 44−26 = 18. Their differences are 4 and 4, so a quadratic rule is appropriate. Test option A: for n = 1, 3(1)² + 1 − 2 = 2; for n = 2, 3(2)² + 2 − 2 = 12; for n = 3, 27 + 3 − 2 = 28, not 26. Thus the supplied key is wrong. Option B gives 6, 12, 22, 36; C gives 2, 12, 22, 32; and D gives 2, 14, 28, 44. None matches all four terms, so the item needs correction.

For \(a_n=7-4n\), \(a_{n+1}-a_n=[7-4(n+1)]-(7-4n)=-4\), which is constant and non-zero for every \(n\). Hence it is an AP. In option B, the difference changes with \(n\). Exam tip: an AP has an nth-term rule linear in \(n\).

Substituting \(n=1,2,3,4\) into \(a_n=2n^3-n\) gives \(1,14,51,124\), respectively. Hence, option B is correct. The close distractor \(n^3+1\) gives 9 when \(n=2\), not the required second term 14. Exam tip: test an explicit rule using at least the first three values of \(n\).

The governing concept is evaluation of an explicit sequence rule followed by simplification of a ratio. Substitute n = 5: a_5 = 5^3 + 5^2 = 125 + 25 = 150. Substitute n = 3: a_3 = 3^3 + 3^2 = 27 + 9 = 36. Therefore a_5 : a_3 = 150 : 36. Dividing both terms by their greatest common divisor, 6, gives 25 : 6. Hence option B is correct. Option A is the unsimplified ratio, option C reverses the order, and option D incorrectly compares only powers.

For \(a_n=n^3+n^2\), substituting \(n=1,2,3,4\) gives \(2,12,36,80\), respectively. Hence, this is the general term of the sequence. The close distractor \(4n^2-2n\) gives 2 and 12 initially, but for \(n=3\) it gives 30, not 36. Exam tip: verify a proposed general term using at least the first three values of \(n\).

(a_3=72) and (a_{10}=30), so the sum is (102). In exams, find both terms carefully in a decreasing formula.

At (n=1) it gives (84), and at (n=2) it gives (78), so (a_n=90-6n). In exams, check the first two terms of a decreasing sequence.

Given \(a_n=\frac{n(2n+3)}{2}\). Substituting \(n=6\), \(a_6=\frac{6(2\times6+3)}{2}=\frac{6\times15}{2}=45\). Hence, 45 is correct. An answer such as 42 may result from evaluating \(2n+3\) incorrectly or making an error in multiplication or division. Exam tip: substitute the value of \(n\) first, then simplify the bracket and calculate step by step.