nth term

For \(a_n=3n^2+2\), substituting \(n=1,2,3,4\) gives \(5,14,29,50\), respectively. Hence, it generates the given sequence. Although \(a_n=n^2+4\) gives the first term as 5, its second term is 8, not 14. Exam tip: test an nth-term formula using at least the first two or three terms.

For the fourth term, substitute \(n=4\): \(a_4=\frac{4(4+3)}{2}=\frac{4\times7}{2}=14\). Therefore, the correct answer is 14. A value such as 16 can result from incorrectly evaluating \(4+3\). Exam tip: in an nth-term formula, substitute the given value of \(n\) first and then simplify step by step.

The differences (3,4,5) are increasing, so the next difference is (6) and the fifth term is (20). In exams, observe changing differences in order.

The governing concept is the general term of an arithmetic sequence. The consecutive differences are 11−7 = 4, 15−11 = 4, and 19−15 = 4, so the common difference is d = 4 and the first term is a₁ = 7. Applying aₙ = a₁ + (n−1)d gives aₙ = 7 + (n−1)4 = 7 + 4n − 4 = 4n + 3. Therefore option A is correct. Substitution verifies the rule: n = 1 gives 7, n = 2 gives 11, n = 3 gives 15, and n = 4 gives 19. Option B has an incorrect coefficient and constant; options C and D also fail to reproduce the stated first term and the constant difference.

Given \(a_n=3n^2-2n+5\), substitute \(n=8\): \(a_8=3(8)^2-2(8)+5=3\times64-16+5=181\). Therefore, 181 is correct. A value such as 189 can result from an error while evaluating \(8^2\) or performing the subtraction. Exam tip: evaluate the power first, then multiply, subtract, and add.

The consecutive differences are \(5,7,9,11\), increasing by 2, so the rule is quadratic. Using \(a_n=n^2+2n-1\) gives \(a_1=2\), \(a_2=7\), \(a_3=14\), and \(a_4=23\). Although \(n^2+1\) gives the first term as 2, it gives the second term as 5, not 7. Exam tip: test an nth-term formula with at least the first two or three terms.

The kth term is given as 47. Substitute k for n in the formula: \(5k-3=47\). Thus, \(5k=50\), so \(k=10\). If 9 is used, the term is \(5\times9-3=42\), not 47. Exam tip: To find a term number, equate the nth-term formula to the given term value and solve for n.

For an arithmetic sequence, \(a_n=a_1+(n-1)d\). The difference between the given terms is \(a_9-a_4=43-18=25\), while the difference between their positions is \(9-4=5\). Hence, \(d=25/5=5\). Using \(a_4=a_1+3d\), we get \(18=a_1+15\), so \(a_1=3\). Therefore, \(a_n=3+(n-1)5=5n-2\). Option \(5n+2\) gives 22 when \(n=4\), so it is incorrect. Exam tip: when two terms are given, first find \(d\) by dividing the difference of the terms by the difference of their indices.

Each term is multiplied by (2), so (a_n=6\cdot2^{n-1}). In a geometric sequence, remember the exponent (n-1).

Given
\(a_n=2n^2+3n-4\). Thus,
\(a_{10}=2(10)^2+3(10)-4=226\) and
\(a_7=2(7)^2+3(7)-4=115\). Therefore,
\(a_{10}-a_7=226-115=111\). A value such as 118 can result from an error while evaluating the squared term. In an exam, calculate the two terms separately before subtracting.

The terms are \(2^2,3^2,4^2,5^2,6^2,\ldots\). Since the first term corresponds to \(n=1\) and is \(2^2\), the \(n\)th term is \((n+1)^2\). The expression \(n^2+1\) gives \(5\) for the second term, whereas the actual second term is \(9\). Exam tip: test a proposed formula by substituting \(n=1\) and \(n=2\).

Given \(a_n=9-2n\), put \(a_n=-31\): \(9-2n=-31\). Thus, \(-2n=-40\), so \(n=20\). Therefore, the 20th term is \(-31\). The 19th term is \(9-2(19)=-29\), so it is not correct. Exam tip: To find a particular term number, substitute the given term value for \(a_n\) and solve for \(n\).

The governing idea is substitution into an explicit nth-term rule. For n = 5, a₅ = 5² + 5 + 2 = 25 + 5 + 2 = 32. For n = 6, a₆ = 6² + 6 + 2 = 36 + 6 + 2 = 44. Adding the two required terms gives a₅ + a₆ = 32 + 44 = 76, not 78. Therefore option B is correct, so the supplied original key needs correction. A quick algebraic check is (25 + 5 + 2) + (36 + 6 + 2) = 61 + 15 = 76. The other options arise from arithmetic or substitution errors.

The first term is (10) and the difference is (-3), so (a_n=10+(n-1)(-3)=13-3n). A decreasing sequence has a negative difference.

Given \(a_n=4n+1\), substitute the complete index \(2m\) for \(n\): \(a_{2m}=4(2m)+1=8m+1\). Hence, option A is correct. \(4m+1\) would be obtained for the index \(m\), not \(2m\). Exam tip: Always substitute a compound index in brackets.

The sequence rule is linear: each term is obtained by multiplying its position by 7 and then subtracting 5. To find the change from one term to the next, first replace n by n+1. This gives \(a_{n+1}=7(n+1)-5=7n+2\). Now subtract the original term: \(a_{n+1}-a_n=(7n+2)-(7n-5)=7\). The variable terms cancel completely.

Therefore, option A, 7, is correct. The result can also be understood without expansion: increasing n by 1 increases \(7n\) by 7, while the constant \(-5\) does not change. Options 5 and 2 confuse the constant or the new expression with the difference. The supplied answer and explanation correctly identify the constant first difference of this linear sequence.

The governing concept is direct evaluation of an explicit sequence rule, with special attention to the parity of n. Substitute n = 6 into aₙ = (−1)ⁿ(2n + 3): a₆ = (−1)⁶(2·6 + 3). Since 6 is even, (−1)⁶ = 1, and 2·6 + 3 = 15. Hence a₆ = 1 × 15 = 15, so option A is correct. If the index had been odd, the factor (−1)ⁿ would be −1 and the result would be negative; this explains the plausible distractor −15. The values ±12 do not result from correctly evaluating 2n + 3 at n = 6.

The consecutive differences are 5, 7, 9, and 11, increasing by 2 each time, so the rule is quadratic. Substituting \(n=1,2,3\) in \(n^2+2n\) gives 3, 8, and 15 respectively; hence the nth term is \(n^2+2n\). The closest distractor, \(n^2+n+1\), gives 7 when \(n=2\), not 8. Exam tip: substitute the first two or three values of \(n\) to verify an nth-term formula quickly.

The nth term is \(a_n=6n-1\). Substituting \(n=15\), we get \(a_{15}=6\times15-1=90-1=89\). Therefore, 89 is correct. The value 91 would result from ignoring the \(-1\) in the formula. Exam tip: for an nth-term question, substitute the required term number for \(n\) in the given formula.

For the term to be 33, set \(2^n+1=33\). Thus, \(2^n=32\). Since \(32=2^5\), we get \(n=5\), so the 5th term is 33. The 4th term is \(2^4+1=17\), so it is not correct. Exam tip: In exponential-term questions, first move the constant to the other side and express the result as a power of the same base.