nth term

Each term increases by (\frac{11}{6}), so (a_n=\frac{11n}{6}). Use a common denominator to see the pattern in fractions.

The terms match (5n^2+1), so none of the listed options is correct. Always test options on the first term.

Given \(a_n=8n^2-7n+4\), substitute \(n+1\) for \(n\): \(a_{n+1}=8(n+1)^2-7(n+1)+4=8n^2+9n+5\). Therefore, \(a_{n+1}-a_n=(8n^2+9n+5)-(8n^2-7n+4)=16n+1\). The option \(16n-7\) results from incorrectly handling the linear term produced when expanding \((n+1)^2\). Exam tip: find \(a_{n+1}\) completely before subtracting \(a_n\).

Given \(a_n=2n^3-3n\), substitute \(n=5\): \(a_5=2(5)^3-3(5)=2\times125-15=250-15=235\). Hence, 235 is correct. 250 is only the value of \(2(5)^3\); the term \(3\times5\) must still be subtracted. Exam tip: evaluate the power first, then multiply and subtract.

The first term is (44) and the difference is (-7), so (a_n=44+(n-1)(-7)=51-7n). Use a negative difference in a decreasing sequence.

For an AP, \(a_n=a_1+(n-1)d\). Substituting \(a_1=11\) and \(d=-4\) gives \(a_n=11-4(n-1)\). In option D, the first term is 7, not 11. Exam tip: always verify the first term first.

The second differences are (4), and (2n^2+10n+6) gives the starting terms. Confirm a quadratic rule using the first three terms.

Given \(a_n=13n-19\), set the nth term equal to 176: \(13n-19=176\). Thus, \(13n=195\), so \(n=15\). Therefore, 176 is the 15th term of the sequence. The 14th term is \(13(14)-19=163\), so it is not correct. Exam tip: To find a term number, equate \(a_n\) to the given term and solve for n.

Each term increases by (\frac{13}{7}), so (a_n=\frac{13n}{7}). In fractions with a common denominator, observe the numerator pattern.

Substituting the values in the given formula gives \(a_9=3(9)^2+5(9)-8=280\) and \(a_6=3(6)^2+5(6)-8=130\). Therefore, \(a_9-a_6=280-130=150\). Option 120 is incorrect because it uses 160 as the value of \(a_6\). Exam tip: Evaluate both terms separately before finding their difference.

The terms are (7\cdot1^2,7\cdot2^2,7\cdot3^2,\ldots), so (a_n=7n^2). Identify the coefficient of the square pattern.

Given a_n=28-5n, a_7=28-5(7)=28-35=-7 and a_{12}=28-5(12)=28-60=-32. Therefore, a_7+a_{12}=-7+(-32)=-39. Option A can result from substituting an incorrect value of n in one of the terms. Exam tip: evaluate each required term separately before adding them.

Write the terms as cubes: \\(64=4^3\\), \\(125=5^3\\), \\(216=6^3\\), and \\(343=7^3\\). The cube bases begin at 4 and increase by 1 for each next term. Therefore, when the term number is \\(n\\), its base is \\(n+3\\): at \\(n=1\\), this gives 4; at \\(n=2\\), it gives 5; and so forth. Thus the nth term is \\(a_n=(n+3)^3\\), so option C is correct.

Substitution verifies the rule: \\(a_1=(1+3)^3=4^3=64\\), \\(a_2=5^3=125\\), \\(a_3=6^3=216\\), and \\(a_4=7^3=343\\). The expression \\(n^3+63\\) has no continuing cube pattern, \\((n+2)^3\\) starts with \\(3^3\\), and \\(4n^3\\) does not produce the listed terms. The consistent starting base and unit increase establish the supplied answer.

The index increases by (6) and the common difference is (10), so the difference is (60). In a linear rule, the coefficient of (n) gives the common difference.

The correct answer is option A: \\(a_n=8\cdot3^{n-1}\\). Each term is three times the preceding term: \\(24/8=3\\), \\(72/24=3\\), and \\(216/72=3\\). Thus the sequence is geometric, with first term \\(a=8\\) and common ratio \\(r=3\\). Using \\(a_n=ar^{n-1}\\), we get \\(a_n=8\cdot3^{n-1}\\). Option A gives 8 at n=1 and reproduces all listed terms. Option B has first term 3, not 8. Option C uses \\(3^n\\), giving 24 for n=1, so the exponent is too large by one. Option D also begins with 24, so it does not describe the given sequence from its first term. Exam cue: test n=1 first; the correct nth-term formula must produce the first listed term.

Given \(a_n=5n^2+2n-9\), \(a_{n+2}=5(n+2)^2+2(n+2)-9=5n^2+22n+15\). Hence, \(a_{n+2}-a_n=(5n^2+22n+15)-(5n^2+2n-9)=20n+24\). The option \(20n+22\) results from an error in calculating the constant term. Exam tip: first expand \((n+2)^2=n^2+4n+4\) carefully.

From (9d=108), (d=12), so (a_{26}=181+9\cdot12=289). It is easier to find a distant term from a nearby given term.

The consecutive differences are \(12-2=10\), \(30-12=18\), \(56-30=26\), and \(90-56=34\). Their second differences are constantly \(8\), so the sequence is quadratic with coefficient \(4\) for \(n^2\). Using \(a_n=4n^2-2n\) gives \(a_1=2\), \(a_2=12\), and \(a_3=30\). Option B gives \(10\) as the second term, so it is not correct. Exam tip: verify a proposed nth-term formula using at least the first three terms.

Given \(a_n=7n-13\). To find \(a_{4n+1}\), substitute the complete index \(4n+1\) for \(n\): \(a_{4n+1}=7(4n+1)-13=28n+7-13=28n-6\). Hence, \(28n-6\) is correct. In \(28n-13\), the effect of the \(+1\) in the index has been omitted. Exam tip: always put a compound subscript in brackets before substituting it into the nth-term formula.

This is an arithmetic progression with first term \(a=23\) and common difference \(d=15\). Its \(n\)th term is \(a_n=a+(n-1)d\). So, \(23+(n-1)\times15=218\) gives \((n-1)\times15=195\), hence \(n-1=13\) and \(n=14\). Therefore, 218 is the 14th term. The 13th term is \(203\), not 218. Exam tip: While finding a term number, remember that the formula contains \(n-1\).

Putting \(n=12\), we get \(a_{12}=\frac{12(12+7)}{4}=\frac{12\times19}{4}=3\times19=57\). Hence, 57 is correct. A value such as 60 can result from incorrectly evaluating \(12+7\) or making an error in multiplication or division. Exam tip: simplify 12 and 4 first, then multiply by 19.

The second differences are (8), and (4n^2+8n-1) gives all starting terms. In a quadratic sequence, test options on the first three terms.

Substituting \(n=6\), \(a_6=9\cdot2^{6-1}+4=9\cdot2^5+4=9\cdot32+4=292\). Hence, the correct answer is 292. The value \(290\) can result from an incorrect addition of the final \(+4\). Exam tip: in exponential expressions, calculate \(n-1\) first and then evaluate the power.

For odd (n), terms are negative and for even (n), terms are positive, so (a_n=4(-1)^n n). In alternating signs, test first at (n=1).

Given \(a_n=n^2+14\), set the term equal to 183: \(n^2+14=183\). Thus, \(n^2=169\), so \(n=13\), since a term number must be a positive integer. Therefore, 183 is the 13th term. For example, the 12th term is \(12^2+14=158\), not 183. Exam tip: To find the position of a given term, equate \(a_n\) to that value and solve for \(n\).