nth term

Given \(a_n=an+b\), we have \(a_2+a_5=(2a+b)+(5a+b)=7a+2b=29\) and \(a_3+a_6=9a+2b=39\). Subtracting the first equation from the second gives \(2a=10\), so \(a=5\). Substituting in \(7a+2b=29\) gives \(b=-3\). Hence, \(a_{10}=10(5)-3=47\). The nearby option 45 is incorrect because it would require \(b=-5\). Exam tip: subtract such paired-sum equations first, as the constant term often cancels out.

The governing concept is forming a general rule for a sequence of fractions by finding separate numerator and denominator patterns. The numerators are 3, 5, 7, 9, which follow 2n + 1 when n starts at 1. The denominators are 2, 3, 4, 5, which follow n + 1. Combining these independent patterns gives aₙ = (2n + 1)/(n + 1). Substitution confirms it: n = 1 gives 3/2, n = 2 gives 5/3, and n = 4 gives 9/5. Option B starts with 1/2, option C reverses the structure, and option D gives 3 rather than 3/2 for n = 1.

The governing concept is evaluating an indexed formula while handling the parity of the exponent. For n = 6, which is even, (−1)⁶ = 1, so a₆ = 6² + 1 = 36 + 1 = 37. For n = 5, which is odd, (−1)⁵ = −1, so a₅ = 5² − 1 = 25 − 1 = 24. Therefore a₆ − a₅ = 37 − 24 = 13. Option A would ignore the sign contribution in one part, option B comes from an arithmetic error, and option D does not result from the stated rule. Separating the even and odd cases prevents the common sign mistake.

The governing concept is recognizing the formula for tetrahedral numbers, or cumulative sums of triangular numbers. The proposed rule in option B gives n = 1: 1·2·3/6 = 1; n = 2: 2·3·4/6 = 4; n = 3: 3·4·5/6 = 10; n = 4: 4·5·6/6 = 20; and n = 5: 5·6·7/6 = 35. Thus it matches every displayed term. The first differences are 3, 6, 10, 15, which are triangular numbers, supporting the same conclusion. Option A is only the triangular-number formula, while options C and D fail when checked against later terms.

For \(a_n=n^2+1\), \(a_{n+1}-a_n=(n+1)^2+1-(n^2+1)=2n+1\). The difference between consecutive first differences is therefore \(2\), which is constant. \(3n-2\) is linear: its first differences are already constant and its second difference is \(0\), so it is not a quadratic function. Exam tip: constant non-zero second differences usually indicate a quadratic sequence.

The governing concept is using finite differences to identify a quadratic nth-term rule. The first differences are 7, 11, 15, and 19; their second differences are 4, 4, and 4. For a quadratic an² + bn + c, the constant second difference is 2a, so 2a = 4 and a = 2. Testing the remaining coefficients gives option B: at n = 1, 2 + 3 + 1 = 6; at n = 2, 8 + 6 + 1 = 15, which does not match the sequence. Therefore the supplied sequence and option B are inconsistent. In fact, solving from the terms gives aₙ = 2n² + n + 3: it yields 6, 13, 24, 39, 58. Thus option A is the unambiguous correct answer; option B is not correct.

We need \(a_n<10\). Thus, \(50-4n<10\), so \(-4n<-40\) and hence \(n>10\). Since \(n\) must be an integer, the smallest possible value is \(11\). In fact, \(a_{10}=10\), which is not less than 10, whereas \(a_{11}=6\). Exam tip: use \(<\), not \(\leq\), for “less than.”

The numerators \(2,5,8,11,\ldots\) form an arithmetic progression with nth term \(2+(n-1)\times3=3n-1\). The denominators \(5,8,11,14,\ldots\) similarly have nth term \(5+(n-1)\times3=3n+2\). Therefore, the nth term of the given sequence is \(\frac{3n-1}{3n+2}\). For example, option D gives \(\frac{2}{4}\) at \(n=1\), not the first term \(\frac{2}{5}\). Exam tip: in fractional sequences, determine the numerator and denominator patterns separately.

Subtracting the two sums gives (2a=12), so (a=6). Then (12a+2b=74) gives (b=1), so (a_1=a+b=7).

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. Since a₇ = a₁ + 6d and d = 6, we get 38 = a₁ + 36, so a₁ = 2. Therefore aₙ = 2 + (n − 1)6 = 2 + 6n − 6 = 6n − 4. Thus option B is correct. A quick check confirms it: substituting n = 7 gives 42 − 4 = 38, exactly as required. Option A has the wrong constant term and gives 44 at n = 7; option C gives 50; option D incorrectly uses 7 as the coefficient of n rather than 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.