nth term

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.