nth term

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 first term is (31) and the difference is (-5), so (a_n=31+(n-1)(-5)=36-5n). Use a negative difference in a decreasing sequence.

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.

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.

Each term is multiplied by (3), so (a_n=5\cdot3^{n-1}). In a geometric sequence, the first term stays separate.

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.

From (9d=63), (d=7), so (a_{23}=95+9\cdot7=158). It is easier to find a distant term from a nearby given term.

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.