nth term

The governing concept is the nth-term formula of an arithmetic sequence. Consecutive terms increase by 7, so the sequence has first term a₁ = 12 and common difference d = 7. Its explicit rule is aₙ = a₁ + (n − 1)d = 12 + 7(n − 1) = 7n + 5. Substituting n = 30 gives a₃₀ = 7(30) + 5 = 210 + 5 = 215. Therefore option B is correct. Option A is 7 less than the correct result and would come from using 29 without the required first-term adjustment; C and D are obtained by adding too much. The explicit formula avoids listing all thirty terms and is the efficient method.

The governing concept is finding an explicit formula for the nth term and verifying it with several initial positions. Test option A: for n = 1, it gives 1(2 + 3)/2 = 5/2; for n = 2, it gives 2(4 + 3)/2 = 7; for n = 3, it gives 3(6 + 3)/2 = 27/2; and for n = 4, it gives 4(8 + 3)/2 = 22. Thus it reproduces every displayed term, so A is correct. Option B gives 2 at n = 1, option C gives 3, and option D gives 2, so those alternatives fail immediately.

The governing concept is verifying a proposed explicit general term by substituting successive positive integers. For option A, n = 1 gives 5 − 2 = 3; n = 2 gives 25 − 4 = 21; n = 3 gives 125 − 6 = 119; and n = 4 gives 625 − 8 = 617. Every displayed term is reproduced, so a_n = 5^n − 2n is correct. Option B grows only linearly, option C gives 3, 5, 9, 17, and option D gives negative values initially or does not match the sequence. Checking several indices prevents accepting a formula that fits only one term.

The governing concept is obtaining an nth-term formula and checking it at each given index, including fractional values. For option A, n = 1 gives (2 + 5)/3 = 7/3; n = 2 gives (8 + 10)/3 = 6; n = 3 gives (18 + 15)/3 = 11; and n = 4 gives (32 + 20)/3 = 52/3. Thus A reproduces the sequence exactly. Option B gives 1, 3, 6, 10, while option C gives 3, 5, 7, 9. Option D gives 2, 13/2, 14, 25/2, so neither matches the listed terms.

The governing concept is recognizing consecutive perfect cubes and expressing their bases in terms of n. The given terms are 64 = 4^3, 125 = 5^3, 216 = 6^3, and 343 = 7^3. The cube bases are therefore 4, 5, 6, 7, which can be written as n + 3 when n = 1, 2, 3, 4. Hence the nth term is a_n = (n + 3)^3, so option B is correct. Option A gives 27 as the first term, option C gives 64 initially but then fails at n = 2, and option D gives 4 only for n = 1 after cubing n, so it does not reproduce the sequence.

The governing concept is finding an explicit rule by identifying a pattern in the terms. Each term is a square: 9 = 3^2, 49 = 7^2, 121 = 11^2, and 225 = 15^2. The bases 3, 7, 11, 15 form an arithmetic sequence with first term 3 and common difference 4. Therefore, the nth base is 3 + 4(n - 1) = 4n - 1, so the nth term is a_n = (4n - 1)^2. Substitution confirms the rule: n = 1 gives 9 and n = 2 gives 49. Option B produces 9, 25, 49, so it does not match; C and D also fail at the second term. Hence option A is correct.

The governing concept is the nth-term formula for an arithmetic sequence. The first term is 15 and the common difference is 9, since 24 - 15 = 9 and 33 - 24 = 9. Therefore, a_n = a_1 + (n - 1)d = 15 + 9(n - 1) = 9n + 6. For the 40th term, substitute n = 40: a_40 = 9(40) + 6 = 360 + 6 = 366. Equivalently, starting at 15 and adding 9 for the 39 intervals before the 40th term gives 15 + 351 = 366. Option A is one step too small, C is 9 too large, and D is 18 too large. Thus option B is correct.

The governing concept is evaluating an explicit rule at successive positive integer values of n. The first term is obtained by putting n = 1: a₁ = 2(1) + 5 = 7. The second term is a₂ = 2(2) + 5 = 9, and the third term is a₃ = 2(3) + 5 = 11. Therefore the first three terms are (7, 9, 11), so option B is correct. The constant increase is 2, confirming that these terms form an arithmetic progression. Option A incorrectly treats 5 as the first term, option C uses a constant pattern beginning at 6, and option D begins at 8; none follows the given formula when n starts at 1.

The governing concept is finding the nth term of an arithmetic progression. Here the first term is a₁ = 13 and the common difference is d = 18 − 13 = 5. The nth-term formula is aₙ = a₁ + (n − 1)d. For the eighth term, a₈ = 13 + (8 − 1)5 = 13 + 35 = 48. Therefore option C is correct. A useful check is to continue the sequence: the fifth term is 33, the sixth is 38, the seventh is 43, and the eighth is 48. Option A is actually the seventh term, so it results from using six differences instead of seven. Options B and D do not follow the fixed difference of five and therefore cannot be the eighth term.

The governing idea is the nth-term rule for an arithmetic progression. The sequence consists of consecutive multiples of 7: 7×1, 7×2, 7×3, and 7×4. Hence its tenth term is 7×10 = 70. Using the formal formula gives the same result: a₁ = 7, d = 7, so a₁₀ = a₁ + (10 − 1)d = 7 + 9×7 = 70. Therefore option B is correct. Option A is the ninth term, 7×9. Option C would be the eleventh term, and option D would be the twelfth term. The position number must be multiplied by 7; simply continuing the pattern without counting the position can cause an off-by-one error.

The governing concept is the nth-term formula aₙ = a₁ + (n − 1)d. In this progression, a₁ = 3 and d = 8 − 3 = 5. For n = 10, a₁₀ = 3 + (10 − 1)×5 = 3 + 45 = 48. Therefore option D is correct. The subtraction of one from the term number is essential because the first term already has its starting value and no difference has yet been added; nine differences are needed to reach the tenth term. Option A results from adding only eight differences, option B from omitting the first term in the calculation, and option C does not satisfy the progression formula.

The governing concept is the general nth-term formula for an arithmetic progression: aₙ = a₁ + (n − 1)d. The first term is a₁ = 12 and the common difference is d = 17 − 12 = 5. Substituting gives aₙ = 12 + (n − 1)5 = 12 + 5n − 5 = 5n + 7. Thus option A is correct. Verification is straightforward: for n = 1, it gives 12; for n = 2, it gives 17; and for n = 3, it gives 22. Option B uses the wrong coefficient and constant, option C gives 17 for n = 1 instead of 12, and option D gives an incorrect difference of 12. The term number must multiply the common difference, with the adjustment based on the first term.

The governing concept is the nth-term formula for an arithmetic progression: a_n = a + (n − 1)d. Here the first term is a = 100 and the common difference is d = 90 − 100 = −10. For n = 11, a_11 = 100 + (11 − 1)(−10) = 100 − 100 = 0. Therefore, option A is correct. The sequence decreases by 10 at every step, so after ten equal gaps it reaches zero. Options B, C, and D do not follow the required ten-step decrease from the first term; they would result from using too few gaps or mishandling the negative common difference.

An arithmetic progression is a sequence in which the same number is added each time. Here the first term is 13, and the common difference is 6 because 19 − 13 = 6. The nth-term rule is \\(a_n=a+(n-1)d\\), where a is the first term and d is the common difference. We must find the position of 61, not merely check whether it appears near the beginning.

Put the given value into the rule: \\(61=13+(n-1)6\\). Subtracting 13 gives \\(48=6(n-1)\\), so \\(n-1=8\\) and \\(n=9\\). Checking the sequence confirms this: the ninth term is 13 plus eight groups of 6, which equals 61. Therefore, option C, the 9th term, is correct. Options 7, 8, and 10 give different terms because they use the wrong number of common differences.

Use the arithmetic-progression rule a_n = a + (n − 1)d. The first term is 11 and the common difference is d = 8 − 11 = −3. Thus the terms continue as 11, 8, 5, 2, −1, −4, ... . The first value below zero is −1, which occurs at n = 5. Therefore the correct answer is option B, the fifth term. The fourth term is 2, which is positive, so it cannot be the first negative term. The sixth and seventh terms are also negative, but they occur later and therefore do not satisfy the word “first.”

In an arithmetic progression, each term is obtained by adding the same common difference. Here d = 13 − 9 = 4, and the terms shown are 9, 13, 17, and 21. The fifth term is therefore 21 + 4 = 25. Equivalently, using a_n = a + (n − 1)d gives a_5 = 9 + (5 − 1)4 = 9 + 16 = 25. Hence option C is correct. The value 23 does not continue the difference of 4, while 24 and 26 arise from incorrect additions or from overlooking that the fourth term is 21 and one more common difference is required.

The sequence has first term a = 2 and common difference d = 8 − 2 = 6. To determine the position of 32, use a_n = a + (n − 1)d. Thus 32 = 2 + (n − 1)6, so 30 = 6(n − 1), n − 1 = 5, and n = 6. Therefore 32 is the sixth term, making option C correct. Listing the terms confirms this: 2, 8, 14, 20, 26, 32. The fourth term is 20, the fifth is 26, and the seventh would be 38, so the other options do not match the progression.

The arithmetic-progression formula is a_n = a + (n − 1)d. The first term is a = 21 and the common difference is d = 27 − 21 = 6. For the eighth term, a_8 = 21 + (8 − 1)6 = 21 + 42 = 63. Hence option C is correct. The seven in the formula is important because there are seven equal gaps from the first term to the eighth term, not eight gaps. Option A corresponds to stopping too early, while 61 and 65 result from using an incorrect difference or making an addition error. The direct continuation also gives 45, 51, 57, 63, confirming the result.

The governing concept is the nth-term formula for an arithmetic progression: a_n = a + (n − 1)d. Here the first term is a = 28 and the common difference is d = 24 − 28 = −4. For n = 9, a9 = 28 + (9 − 1)(−4) = 28 + 8(−4) = 28 − 32 = −4. Hence option A is correct. The negative answer is reasonable because the sequence decreases by 4 at every step: the terms are 28, 24, 20, 16, 12, 8, 4, 0, −4. Options B, C, and D result from using the wrong number of steps or stopping before the ninth term. The formula counts eight differences from the first to the ninth term.

The governing concept is the nth-term formula for an arithmetic progression: aₙ = a + (n − 1)d. The fifth term is given as a₅ = 27 and the common difference is d = 3. Substituting n = 5 gives 27 = a + (5 − 1)3 = a + 12. Therefore, a = 27 − 12 = 15, so option B is correct. Another way to see it is to move backward four equal steps from the fifth term: 27, 24, 21, 18, 15. Option A subtracts five differences instead of four, while options C and D subtract only three or two differences. Since the first term is four positions before the fifth term, exactly four common differences must be removed.