nth term

Correct answer: B. -30

Explanation: Given \(a_n=17-4n\), \(a_5=17-4(5)=-3\) and \(a_{11}=17-4(11)=-27\). Therefore, \(a_5+a_{11}=-3+(-27)=-30\). A value such as \(-28\) can result from an error while adding the negative terms. Exam tip: find each required term separately and then add them with their signs.

Correct answer: C. ((n+2)^3)

Explanation: The sequence can be recognized by writing each term as a cube: \\(27=3^3\\), \\(64=4^3\\), \\(125=5^3\\), and \\(216=6^3\\). The bases increase by 1, so the first term corresponds to base 3. If the term number is \\(n\\), its base is therefore \\(n+2\\): for \\(n=1\\), the base is 3; for \\(n=2\\), it is 4; and so on. Hence the general term is \\(a_n=(n+2)^3\\), making option C correct.

Checking the formula confirms it: \\(a_1=(1+2)^3=27\\), \\(a_2=4^3=64\\), \\(a_3=5^3=125\\), and \\(a_4=6^3=216\\). The expression \\(n^3+26\\) matches the first term but not the later pattern, while \\((n+1)^3\\) starts at \\(2^3\\), and \\(3n^3\\) gives a different sequence. Thus, the supplied answer follows directly from identifying consecutive cubes.

Correct answer: D. (40)

Explanation: The index increases by (5) and the common difference is (8), so the difference is (40). In a linear rule, the coefficient of (n) gives the common difference.

Correct answer: A. (5\cdot3^{n-1})

Explanation: The correct answer is option A: \\(a_n=5\cdot3^{n-1}\\). The sequence is geometric because every term is multiplied by 3: \\(15/5=3\\), \\(45/15=3\\), and \\(135/45=3\\). The nth-term rule is \\(a_n=ar^{n-1}\\). Substituting the first term \\(a=5\\) and ratio \\(r=3\\) gives \\(a_n=5\cdot3^{n-1}\\). Option A gives \\(5\\) when \\(n=1\\), \\(15\\) when \\(n=2\\), and \\(45\\) when \\(n=3\\), so it matches. Option B starts with 3 rather than 5. Option C has exponent n, so its first value is 15 instead of 5. Option D starts with 15 and therefore represents a shifted sequence, not the given one. Common mistake: do not use the ratio as the first term; keep the first term outside the power.

Correct answer: C. \(16n+18\)

Explanation: Given \(a_n=4n^2+n-6\), substitute \(n+2\) for \(n\): \(a_{n+2}=4(n+2)^2+(n+2)-6=4n^2+17n+12\). Hence, \(a_{n+2}-a_n=(4n^2+17n+12)-(4n^2+n-6)=16n+18\). Option B results from an incorrect expansion or simplification of the constant terms. Exam tip: replace every occurrence of \(n\) with \(n+2\) before expanding the expression.

Correct answer: C. (158)

Explanation: The correct answer is option C: \\(158\\). In an arithmetic sequence, consecutive terms differ by the same common difference \\(d\\). From the fifth term to the fourteenth term there are \\(14-5=9\\) equal steps. The change is \\(95-32=63\\), so \\(9d=63\\) and \\(d=7\\). From the fourteenth term to the twenty-third term there are \\(23-14=9\\) steps again. Therefore \\(a_{23}=a_{14}+9d=95+9(7)=95+63=158\\). Option A, 154, would use a change of 59 and is too small. Option B, 156, is also two less than the correct result. Option C exactly follows nine increases of 7. Option D, 160, is two too large. A useful cue is that equal index gaps in an arithmetic sequence produce equal value changes.

Correct answer: A. \(2n^2-1\)

Explanation: The consecutive differences are \(6,10,14,18\), and their second differences are constantly \(4\). Hence, the nth term is quadratic in \(n\). Substituting \(n=1,2,3\) in \(2n^2-1\) gives \(1,7,17\), respectively, so it matches the sequence. The close distractor \(2n^2+n-2\) gives \(8\) when \(n=2\), not the required second term \(7\). Exam tip: a constant second difference usually indicates a quadratic nth-term rule.

Correct answer: B. \(18n-20\)

Explanation: Given \(a_n=6n-8\), substitute the entire index \(3n-2\) for \(n\): \(a_{3n-2}=6(3n-2)-8=18n-12-8=18n-20\). Hence, \(18n-20\) is correct. The option \(18n-12\) results from forgetting to subtract the final 8. Exam tip: always enclose a compound index in brackets before simplifying.

Correct answer: B. 12th

Explanation: This is an arithmetic progression with first term \(a=19\) and common difference \(d=12\). Its \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(19+(n-1)\times12=151\), so \((n-1)\times12=132\), giving \(n-1=11\) and \(n=12\). Hence, 151 is the 12th term. The 13th term would be 163, so it is not correct. Exam tip: while finding a term number, use \(n-1\) correctly in the AP formula.

Correct answer: C. 68

Explanation: 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.

Correct answer: B. (3n^2+4n+1)

Explanation: 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.

Correct answer: C. 109

Explanation: 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.

Correct answer: B. (3(-1)^{n+1}n)

Explanation: 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).

Correct answer: B. 11th

Explanation: 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.

Correct answer: C. \(24\left(\frac{1}{2}\right)^{n-1}\)

Explanation: 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).

Correct answer: C. 301

Explanation: 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.

Correct answer: A. (426)

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

Correct answer: C. 111

Explanation: 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.

Correct answer: A. (6n^2)

Explanation: 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.

Correct answer: B. 5th term

Explanation: 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 \(n\).

Correct answer: B. (3n^2+7n+2)

Explanation: 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).

Correct answer: A. \(20p+24\)

Explanation: 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 \(n\).

Correct answer: A. 2ⁿ + 3

Explanation: 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.

Correct answer: B. 7th

Explanation: 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.

Correct answer: B. (12)

Explanation: 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.