nth term

The governing concept is identifying an explicit formula by testing the position number n against the observed terms. For option C, when n = 1, n² + 3n = 1 + 3 = 4. When n = 2, it gives 4 + 6 = 10; when n = 3, it gives 9 + 9 = 18; when n = 4, it gives 16 + 12 = 28; and when n = 5, it gives 25 + 15 = 40. Thus aₙ = n² + 3n reproduces every listed term, so option C is correct. Option A gives 3 for the first term, option B gives 20 for the third term, and option D gives 5 for the first term. Checking several positions makes the choice reliable rather than based on only one coincidence.

The governing concept is finding a general rule for a sequence of fractions by examining the numerator and denominator separately. The numerators are 3, 5, 7, and 9. Starting with n = 1, this pattern is 2n + 1. The denominators are 2, 3, 4, and 5, which follow the rule n + 1. Combining the two independent patterns gives aₙ = (2n + 1)⁄(n + 1), so option A is correct. Verification is immediate: n = 1 gives 3⁄2, n = 2 gives 5⁄3, and n = 4 gives 9⁄5. Option B begins with 1⁄2, option C reverses the roles and structure of the expressions, and option D gives 3 when n = 1 rather than 3⁄2.

The governing concept is evaluating an indexed formula while correctly handling the parity of the exponent. Since 6 is even, (−1)⁶ = 1, so a₆ = 6² + 1 = 36 + 1 = 37. Since 5 is odd, (−1)⁵ = −1, so a₅ = 5² − 1 = 25 − 1 = 24. Therefore a₆ − a₅ = 37 − 24 = 13, making option C correct. The sign of (−1)ⁿ must be evaluated separately for each index; treating both powers as positive would produce an incorrect result. Option A can arise from omitting one sign contribution, option B from an arithmetic error, and option D does not follow from the stated formula. The even–odd check confirms the answer.

The first term is (14) and the difference is (8), so (a_n=14+(n-1)8=8n+6). In an arithmetic sequence, (n-1) differences are added.

Substitute \(n=6\): \(a_6=5(6)^2-4(6)+3=5\times36-24+3=159\). Hence, the correct value is 159. The value 153 may result from omitting the constant term 3. Exam tip: evaluate the power first, then perform multiplication and addition/subtraction.

The terms are 7, 18, 33, 52, and 75. Their first differences are 11, 15, 19, and 23, which increase by 4 each time. This indicates a quadratic formula. Check option B, \\(a_n=2n^2+5n\\): at n = 1 it gives \\(2+5=7\\); at n = 2 it gives \\(8+10=18\\); at n = 3 it gives \\(18+15=33\\); and at n = 4 it gives \\(32+20=52\\). It also gives 75 at n = 5, so B is correct.

The constant second difference is 4. For a quadratic expression \\(an^2+bn+c\\), the second difference equals \\(2a\\), so here the coefficient of \\(n^2\\) is 2. Since the first term is 7, the remaining linear part is fixed by the next terms, giving \\(2n^2+5n\\). Option A fails at the first term, while the other choices also fail on direct substitution. The supplied answer B is accurate.

The governing concept is substitution of a composite index into an explicit sequence formula. The rule is aₙ = 9n − 11, and the requested index is 2k + 3. Replace the entire n by 2k + 3: a₂ₖ₊₃ = 9(2k + 3) − 11. Distributing 9 gives 18k + 27 − 11, which simplifies to 18k + 16. Therefore option A is correct. Option B stops before subtracting 11, option C incorrectly multiplies only k by 9, and option D substitutes the expression incompletely. Keeping the complete index inside parentheses is essential to avoid these errors.

Given \(a_n=n^2+7n-8\), \(a_9=9^2+7(9)-8=81+63-8=136\) and \(a_4=4^2+7(4)-8=16+28-8=36\). Hence, \(a_9-a_4=136-36=100\). A value such as 104 can result from substituting an incorrect value for one of the terms. Exam tip: calculate each required term separately before finding their difference.

The successive differences are \(8,12,16,20\), and their second differences are all \(4\). Hence, the nth term has a quadratic form. Substituting \(n=1,2,3\) in \(a_n=2n^2+2n-1\) gives \(3,11,23\), so this is the correct rule. The close distractor \(2n^2+n\) gives the first term as 3 but gives 10, not 11, when \(n=2\). Exam tip: equal second differences usually indicate a rule of the form \(an^2+bn+c\).

Given \(a_n=26-5n\). Put \(a_n=-34\): \(26-5n=-34\). Thus, \(-5n=-60\), so \(n=12\). Therefore, the 12th term is \(-34\). The 11th term is \(-29\), so it is not correct. Exam tip: when solving equations involving negative numbers, track the signs carefully.

The terms match (n^2+5n), giving (6) for (n=1) and (14) for (n=2). Test options on the starting terms.

For (n=8), ((-1)^9=-1) and (4(8)+1=33), so (a_8=-33). In alternating signs, check whether the exponent is even or odd.

The governing concept is the nth-term formula for an arithmetic progression: aₙ = a₁ + (n − 1)d. For the seventh term, a₇ = a₁ + 6d. Since a₇ = 38 and d = 6, we have 38 = a₁ + 6(6) = a₁ + 36, giving a₁ = 2. Substitute this into the general formula: aₙ = 2 + (n − 1)6 = 2 + 6n − 6 = 6n − 4. Therefore option B is correct. A direct check gives a₇ = 6(7) − 4 = 42 − 4 = 38. Option A gives 44 at n = 7, option C gives 50, and option D incorrectly uses 7 as the coefficient instead of the common difference 6.

The sequence is not arithmetic because its successive differences are 8, 16, 32, and 64. Instead, each term can be written as a power of 2 plus 2: 10 = 2³ + 2, 18 = 2⁴ + 2, 34 = 2⁵ + 2, 66 = 2⁶ + 2, and 130 = 2⁷ + 2. The exponent is n + 2 when the first term corresponds to n = 1. Hence the general term is aₙ = 2ⁿ⁺² + 2, so option A is correct. Option B gives 10 for n = 1 but then gives 22 for n = 2, not 18. Option C gives 12 as the first term, and option D describes a linear pattern, which this sequence clearly is not.

Each term increases by (\frac{7}{4}), so (a_n=\frac{7n}{4}). Use a common denominator to see the pattern in fractions.

The sequence is 4, 13, 28, 49, 76, and the first differences are 9, 15, 21, and 27. These increase by 6, so the second difference is constant and the nth-term formula is quadratic. Test option B, \\(a_n=3n^2+1\\): for n = 1, it gives \\(3(1)^2+1=4\\); for n = 2, it gives \\(3(2)^2+1=13\\); and for n = 3, it gives \\(3(3)^2+1=28\\). At n = 4 and 5 it gives 49 and 76. Thus option B is correct.

For a quadratic expression \\(an^2+bn+c\\), the second difference is \\(2a\\). Since the second difference here is 6, the coefficient of \\(n^2\\) must be 3. The constant term is then fixed by the first term: \\(3(1)^2+c=4\\), so \\(c=1\\). The formula becomes \\(3n^2+1\\). The other choices do not reproduce the listed terms. Therefore the supplied answer B is mathematically sound.

Given \(a_n=6n^2-5n+2\), substitute \(n+1\) for \(n\): \(a_{n+1}=6(n+1)^2-5(n+1)+2=6n^2+7n+3\). Hence, \(a_{n+1}-a_n=(6n^2+7n+3)-(6n^2-5n+2)=12n+1\). Therefore, option B is correct. \(12n-5\) can result from an error while subtracting the constant terms. Exam tip: expand \((n+1)^2\) as \(n^2+2n+1\) before simplifying.

Given \(a_n=n^3+2n\), substitute \(n=5\): \(a_5=5^3+2(5)=125+10=135\). Hence, 135 is correct. A value such as 130 results from an incorrect evaluation of the linear term \(2n\). Exam tip: after substituting the value of \(n\), evaluate powers and multiplication before addition.

The direct answer is option B: \(a_n=36-5n\). The sequence is arithmetic because each term decreases by 5: 26−31 = −5, 21−26 = −5, and so forth. Use \(a_n=a_1+(n-1)d\). With first term \(a_1=31\) and common difference d = −5, \(a_n=31+(n-1)(-5)=31-5n+5=36-5n\). Option A, 31−5n, is wrong because at n = 1 it gives 26 instead of 31. Option B is correct because it gives 31 for n = 1, 26 for n = 2, 21 for n = 3, and continues exactly as required. Option C, 5n+26, increases by 5 and gives 31 only at n = 1; its next term is 36, so it does not match. Option D, 35−4n, decreases by 4 and gives 31 first, but then 27 rather than 26. The common mistake is forgetting the \((n-1)\) in the arithmetic-progression formula. Check the first term and the difference before choosing.

Given \(a_n=3^n+2n\), substitute \(n=4\): \(a_4=3^4+2(4)=81+8=89\). Hence, 89 is the correct answer. The value 87 may result from incorrectly taking \(2n\) as 6. Exam tip: Substitute the term number carefully in every part containing \(n\).

The consecutive differences are 12, 14, 16, and 18; their second differences are 2, so the nth term is quadratic. For \(a_n=n^2+9n+5\), we get \(a_1=15\), \(a_2=27\), and \(a_3=41\). Hence, \(n^2+9n+5\) is correct. The close distractor \(n^2+8n+6\) gives 15 at \(n=1\), but gives 26 at \(n=2\), not 27. Exam tip: for a quadratic sequence, check both first and second differences.

Given \(a_n=11n-17\), set the term equal to 159: \(11n-17=159\). Thus, \(11n=176\), so \(n=16\). Therefore, 159 is the 16th term of the sequence. The 15th term is \(148\), so it is not correct. Exam tip: To find a term number, substitute the given term value for \(a_n\) and solve for \(n\).

Each term increases by (\frac{9}{5}), so (a_n=\frac{9n}{5}). In fractions with a common denominator, observe the numerator pattern.

Given \(a_n=2n^2+3n-4\), \(a_8=2(8)^2+3(8)-4=128+24-4=148\) and \(a_5=2(5)^2+3(5)-4=50+15-4=61\). Therefore, \(a_8-a_5=148-61=87\). Option 84 may seem close, but it does not result from evaluating both terms correctly. Exam tip: calculate each required term separately before finding their difference.

The terms are (3\cdot1^2,3\cdot2^2,3\cdot3^2,\ldots), so (a_n=3n^2). Recognize the square pattern with its coefficient.