nth term

To obtain the value 127, set \(5n^2+2=127\). This gives \(5n^2=125\), so \(n^2=25\) and \(n=5\). Since a term number is a positive integer, the correct answer is the 5th term. The 4th term is \(5\times4^2+2=82\), not 127. Exam tip: equate the given value to \(a_n\) first, then solve for \(n\).

Each term is multiplied by \(\frac{1}{2}\), so \(a_n=12\left(\frac{1}{2}\right)^{n-1}\). Keep the first term separate in a geometric sequence.

Given \(a_n=2n^2-3n+4\), replace \(n\) by \(n+1\): \(a_{n+1}=2(n+1)^2-3(n+1)+4=2n^2+n+3\). Therefore, \(a_{n+1}-a_n=(2n^2+n+3)-(2n^2-3n+4)=4n-1\). The option \(4n+1\) can result from an error while subtracting the constant terms. Exam tip: write \(a_{n+1}\) separately before subtracting \(a_n\).

Observe the terms in order: \(0=1^2-1\), \(3=2^2-1\), \(8=3^2-1\), \(15=4^2-1\), and \(24=5^2-1\). Therefore, the \(n\)th term is \(a_n=n^2-1\). For \(n^2+n\), the first term would be 2, so it does not match this sequence. Exam tip: substitute \(n=1\) and \(n=2\) to verify a proposed nth-term formula quickly.

To find the eighth term, substitute n=8: a_8=3(8)^2+8-2=3×64+8-2=192+8-2=198. Hence, 198 is correct. A value of 202 may result from incorrectly adding the final −2. Exam tip: evaluate the power first, then multiply, add, and subtract in order.

The terms match (1^2+4(1)+1,2^2+4(2)+1,\ldots). Testing options on the first three terms is a good method.

The first term is (9) and the difference is (7), so (a_n=9+(n-1)7=7n+2). In an arithmetic sequence, (n-1) differences are added.

To find the seventh term, substitute n=7: a_7=4(7)^2-3(7)+2=4×49-21+2=196-21+2=177. Therefore, 177 is the correct answer. A value such as 181 usually results from an error in subtraction or addition. Exam tip: evaluate the power first, then perform multiplication and addition/subtraction.

Given \(a_n=7n-4\), substitute \(3m\) for \(n\): \(a_{3m}=7(3m)-4=21m-4\). Hence, the correct answer is \(21m-4\). The expression \(21m-12\) incorrectly multiplies the constant \(-4\) by 3. Exam tip: when the index changes, replace only \(n\) throughout the formula and do not alter a constant term unnecessarily.

(a_3=4) and (a_8=34), so the correct sum is (38). Recheck calculation if options seem close.

The second differences are (4), and the terms match (2n^2+n+1). Equal second differences suggest a quadratic rule.

Given \(a_n=18-4n\), set the nth term equal to \(-30\): \(18-4n=-30\). Thus, \(-4n=-48\), so \(n=12\). Checking, \(a_{12}=18-4(12)=18-48=-30\). Hence, the 12th term is correct. The 13th term gives \(-34\), not \(-30\). Exam tip: handle signs carefully when solving equations involving negative numbers.

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.

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 direct answer is option B: \(a_n=23-3n\). This is an arithmetic progression because the difference between consecutive terms is constant: 17−20 = −3, 14−17 = −3, and so on. The nth-term formula is \(a_n=a_1+(n-1)d\). Here \(a_1=20\) and d = −3, so \(a_n=20+(n-1)(-3)=20-3n+3=23-3n\). Option A, 20−3n, is wrong because for n = 1 it gives 17, not the first term 20. Option B is correct because n = 1 gives 20, n = 2 gives 17, and n = 5 gives 8, matching the sequence. Option C, 3n+17, increases as n increases and gives 20 first but then 23, not 17. Option D, 21−2n, has common difference −2 rather than −3 and gives 19 for n = 1. Exam cue: for a decreasing arithmetic progression, use a negative common difference and test the formula at n = 1.

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.