nth term

Substitute \(n=12\) into \(a_n=\frac{n(n+5)}{3}\): \(a_{12}=\frac{12(12+5)}{3}=\frac{12\times17}{3}=4\times17=68\). Hence, 68 is correct. The distractor 66 can result from incorrectly adding \(12+5\). In an exam, substitute the term number first and then simplify systematically.

The second differences are (6), and (3n^2+4n+1) gives all starting terms. In a quadratic sequence, test options on the first three terms.

Substituting \(n=5\) gives \(a_5=7\cdot2^{5-1}-3=7\cdot2^4-3=7\cdot16-3=112-3=109\). Hence, 109 is correct. Option 111 may result from forgetting to subtract 3 at the end. Exam tip: in expressions with exponents, calculate \(n-1\) first and then evaluate the power.

For odd (n), terms are positive and for even (n), terms are negative, so (a_n=3(-1)^{n+1}n). In alternating signs, test first at (n=1).

To find which term has value 131, set \(a_n=131\): \(n^2+10=131\). Thus, \(n^2=121\), so \(n=11\). Since a term number is a positive integer, 131 is the 11th term of the sequence. The 12th term is \(12^2+10=154\), so it is not correct. Exam tip: To find the position of a given term, equate \(a_n\) to the given value first.

Each term is halved, so \(a_n=24\left(\frac{1}{2}\right)^{n-1}\). In a geometric sequence, the exponent starts with (n-1).

Given \(a_n=4n^2-9n+7\), \(a_{10}=4(10)^2-9(10)+7=317\) and \(a_3=4(3)^2-9(3)+7=16\). Therefore, \(a_{10}-a_3=317-16=301\). A nearby value such as 299 can result from an error while evaluating \(a_3\) or subtracting. Exam tip: substitute each value of \(n\) separately before finding the difference.

The first term is (18) and the difference is (17), so (a_{25}=18+24\cdot17=426). The (25)th term includes (24) differences.

Given \(a_n=3n^2+4n\), \(a_7=3(7)^2+4(7)=147+28=175\) and \(a_4=3(4)^2+4(4)=48+16=64\). Therefore, \(a_7-a_4=175-64=111\). An answer such as 108 can result from an error while evaluating \(7^2\) or \(4^2\). Exam tip: calculate the two terms separately before subtracting.

The terms are (6\cdot1^2,6\cdot2^2,6\cdot3^2,\ldots), so (a_n=6n^2). Identify the coefficient of the square pattern.

Given \(a_n=40-7n\). For \(a_n=5\), set \(40-7n=5\). Thus, \(7n=35\), so \(n=5\). Hence, 5 is the 5th term. The closest distractor, the 4th term, is incorrect because \(a_4=40-28=12\), not 5. Exam tip: To find the position of a specified term, substitute its value in the formula for \(a_n\) and solve for \(n\).

The second differences are (6), and (3n^2+7n+2) gives the starting terms. In a quadratic rule, half of the second difference is the coefficient of (n^2).

Given \(a_n=5n+14\), substitute the complete index \(4p+2\) for \(n\): \(a_{4p+2}=5(4p+2)+14=20p+10+14=20p+24\). Therefore, \(20p+24\) is correct. The option \(20p+14\) misses the contribution \(5\times2=10\). Exam tip: when an index is an expression, substitute the entire expression in brackets for \(n\).

The terms follow a doubling pattern with a fixed addition. They can be rewritten as 5 = 2¹ + 3, 7 = 2² + 3, 11 = 2³ + 3, 19 = 2⁴ + 3, and 35 = 2⁵ + 3. Therefore, when the first term is indexed by n = 1, the general term is aₙ = 2ⁿ + 3. Option A is correct. Substitution verifies every listed term, not merely the first one. Option B gives 5 for n = 1 but gives 9 for n = 2, so its exponent or constant is unsuitable. Option C produces 5, 6, 11, 19, 35 only inconsistently, and option D is linear, with constant first differences, unlike this sequence.

Given \(a_n=n(n+4)\), set \(n(n+4)=77\). This gives \(n^2+4n-77=0\), or \((n-7)(n+11)=0\). Since a term number must be positive, \(n=7\). Therefore, 77 is the 7th term. Although \(n=-11\) is an algebraic root, it cannot be a term number. Exam tip: for sequence indices, accept only positive integer values.

In (a_n=12n-7), the coefficient of (n) is (12), so the common difference is (12). In a linear (n)th term, the coefficient of (n) gives the difference.

Each term is multiplied by \(\frac{1}{3}\), so \(a_n=81\left(\frac{1}{3}\right)^{n-1}\). Keep the first term fixed and use exponent (n-1).

The first term is (17) and the difference is (8), so (a_n=17+(n-1)8=8n+9). In an arithmetic sequence, (n-1) differences are added.

Substitute n=6: a_6=7(6)^2-3(6)+5=7×36-18+5=252-18+5=239. Therefore, 239 is correct. Option 252 is only the value of 7×6² and ignores the terms -3n+5. Exam tip: evaluate the exponent first, then perform multiplication and addition or subtraction.

For an arithmetic progression, the difference between terms equals the number of steps multiplied by the common difference. From a₁₃ − a₆ = (13 − 6)d, we obtain 110 − 47 = 7d, so 63 = 7d and d = 9. Now use a₆ = a₁ + 5d: 47 = a₁ + 45, giving a₁ = 2. Hence aₙ = a₁ + (n − 1)d = 2 + 9(n − 1) = 9n − 7, so option B is correct. Checking gives a₆ = 54 − 7 = 47 and a₁₃ = 117 − 7 = 110. The other options fail one or both checks.