nth term

The terms are generated by (4n^2-2n) only if checked carefully against the sequence. Always test more than one term.

Substitute n=5 directly into the given formula. The sign factor becomes \((-1)^{5+1}=(-1)^6=1\), because an even power of -1 equals 1. The numerical factor is \(3(5)-2=15-2=13\). Multiplying these parts gives \(a_5=1\times13=13\).

Therefore option A is correct. The most important point is to evaluate both parts of the product and to check the parity of the exponent before assigning the sign. If the exponent had been odd, the sign would have been negative, but here it is even. The value \(-13\) would result from missing this sign check, while 17 and -17 come from an arithmetic error in \(3n-2\). The supplied answer and explanation are correct.

The governing concept is the arithmetic-progression formula aₙ = a₁ + (n − 1)d. Since a₈ = a₁ + 7d, substitute the given values: 41 = a₁ + 7 × 5 = a₁ + 35. Thus a₁ = 6. The general term is then aₙ = 6 + (n − 1)5 = 6 + 5n − 5 = 5n + 1. Therefore option B is correct, so the supplied answer A needs correction. Verification gives a₈ = 5 × 8 + 1 = 41. Option A gives 46 at n = 8, while options C and D give 36 and 75 respectively, so they fail the given condition.

Substituting \(n=11\) gives \(a_{11}=\frac{3(11)+7}{2}=\frac{33+7}{2}=\frac{40}{2}=20\). Hence, the correct answer is 20. Although 21 is a close distractor, it does not result from the correct calculation. Exam tip: After substituting the value of \(n\), simplify the numerator first and then divide by 2.

The successive differences are \(5,9,13,17\), and their differences are constantly \(4\). Hence the nth term should be a quadratic expression in \(n\). With \(a_n=n(2n-1)=2n^2-n\), we get \(a_1=1\), \(a_2=6\), \(a_3=15\), \(a_4=28\), and \(a_5=45\). Therefore, option D is correct. Option C gives \(3\) when \(n=1\), so it cannot represent the sequence. Exam tip: a constant second difference usually indicates a quadratic nth-term formula.

Given \(a_n=5n^2-2n+1\), substitute \(n+1\) for \(n\): \(a_{n+1}=5(n+1)^2-2(n+1)+1=5n^2+8n+4\). Therefore, \(a_{n+1}-a_n=(5n^2+8n+4)-(5n^2-2n+1)=10n+3\). The distractor \(10n-2\) results from missing the \(10n\) term produced while expanding \((n+1)^2\). Exam tip: write \(a_{n+1}\) separately before subtracting \(a_n\).

Given a_n=2n^3+n, substitute n=4: a_4=2(4)^3+4=2×64+4=132. Therefore, 132 is the correct option. The value 128 represents only 2×4^3 and misses the final +4 term. Exam tip: after substituting n, evaluate the power first, then multiply and add.

The first term is (20) and the difference is (-3), so (a_n=20+(n-1)(-3)=23-3n). Take the difference as negative in a decreasing sequence.

The terms match (n^2+5n+4), giving (10) for (n=1) and (18) for (n=2). Always test the option on the first two terms.

Given \(a_n=9n-13\) and the term value is \(122\), set \(9n-13=122\). Thus, \(9n=135\), so \(n=15\). Hence, \(122\) is the 15th term of the sequence. The close distractor, the 14th term, is incorrect because \(a_{14}=9(14)-13=113\). Exam tip: To find the position of a given term, equate \(a_n\) to that term’s value.

Each term increases by (\frac{5}{3}), so (a_n=\frac{5n}{3}). Identify the pattern by using a common denominator.

Given \(a_n=n^2+4n-6\), \(a_{10}=10^2+4(10)-6=134\) and \(a_6=6^2+4(6)-6=54\). Therefore, \(a_{10}-a_6=134-54=80\). The value 74 can result from an arithmetic error while evaluating \(a_6\) or subtracting. Exam tip: calculate each required term separately before finding their difference.

The terms are (2\cdot1^2,2\cdot2^2,2\cdot3^2,\ldots), so (a_n=2n^2). Recognize the square pattern with its coefficient.

(a_4=-1) and (a_9=-16), so the sum is (-17). Keep signs in mind when adding negative numbers.

This is a cube sequence and (512=8^3), so it is the (8)th term. In a cube sequence, the term number matches the cube root.

The index increases by (4) and the common difference is (6), so the difference is (24). In a linear rule, the coefficient is the common difference.

Each term is multiplied by (4), so (a_n=6\cdot4^{n-1}). In a geometric sequence, the first term stays separate.

Given \(a_n=3n^2+2n-5\), we get \(a_{n+2}=3(n+2)^2+2(n+2)-5=3n^2+14n+11\). Hence, \(a_{n+2}-a_n=(3n^2+14n+11)-(3n^2+2n-5)=12n+16\). The option \(12n+20\) results from an incorrect subtraction of the constant terms. Exam tip: substitute \(n+2\) into every term first, then combine like terms carefully.

From (9d=45), (d=5), so (a_{20}=59+8\cdot5=99). It is easier to find a far term from a nearby given term.

The terms should be matched carefully with the option. Test all starting terms before finalizing the rule.