nth term

The governing concept is locating a specified value in an arithmetic progression. Use aₙ = a + (n − 1)d. Here the first term is a = 90 and the common difference is d = 82 − 90 = −8. Set the nth term equal to 18: 18 = 90 + (n − 1)(−8). Subtracting 90 gives −72 = −8(n − 1). Dividing by −8 gives n − 1 = 9, so n = 10. Therefore, 18 is the tenth term and option B is correct. A check by listing terms gives 90, 82, 74, 66, 58, 50, 42, 34, 26, 18. The negative difference must be retained; ignoring it or counting from zero can produce the distractor positions.

The governing concept is the nth-term rule for an arithmetic progression: aₙ = a₁ + (n − 1)d. The first term is a₁ = 2, and the common difference is d = 9 − 2 = 7. Therefore, a₈ = 2 + (8 − 1)×7 = 2 + 49 = 51, while a₃ = 2 + (3 − 1)×7 = 2 + 14 = 16. Adding the requested terms gives a₈ + a₃ = 51 + 16 = 67, so option A is correct. The factor n − 1 is essential because the first term is reached after zero common differences. Using n instead of n − 1, choosing the wrong indexed terms, or making an addition error can produce distractors such as 69, 71, or 73. The sequence itself confirms the calculation: its terms are 2, 9, 16, 23, 30, 37, 44, 51, so the eighth and third terms are indeed 51 and 16.

For an arithmetic progression, the nth term is found using \(a_n=a+(n-1)d\), where \(a\) is the first term and \(d\) is the common difference. In the given sequence, the first term is \(a=11\). The difference is \(18-11=7\), and the later differences also equal 7. We need the twelfth term, so use \(n=12\).

Substitution gives \(a_{12}=11+(12-1)7=11+11\times7=11+77=88\). Hence the twelfth term is 88, which is choice C. A common error is to multiply 12 by 7 without subtracting 1; that would count one extra difference. The first term itself uses zero differences, so the twelfth term uses exactly eleven differences.

The governing concept is an explicit nth-term rule: the formula gives the value of a term directly from its position n. To find which position has value 96, set 7n − 2 = 96. Adding 2 to both sides gives 7n = 98, and dividing by 7 gives n = 14. Thus 96 is the fourteenth term. Substitution verifies the result: a₁₄ = 7(14) − 2 = 98 − 2 = 96. The distractors can be checked directly: a₁₁ = 77 − 2 = 75, a₁₂ = 84 − 2 = 82, and a₁₃ = 91 − 2 = 89. None equals 96, so option D is the only correct answer. The calculation also confirms that the requested position is a valid positive integer.

The governing concept is the nth-term formula aₙ = a₁ + (n − 1)d. The first term is 50, and the common difference is 43 − 50 = −7 because each term decreases by 7. Set the nth term equal to 1: 50 + (n − 1)(−7) = 1. Hence 50 − 7(n − 1) = 1, so 7(n − 1) = 49, n − 1 = 7, and n = 8. Verification gives a₈ = 50 + 7(−7) = 1. The nearby terms are a₇ = 8 and a₉ = −6, proving that the seventh and ninth choices are not correct. Therefore option B is the unique answer. The negative common difference must be retained; treating it as positive would produce an incorrect position.

The governing concept is the nth-term formula for an arithmetic progression: aₙ = a₁ + (n − 1)d. The first term is a₁ = 5, and the common difference is d = 14 − 5 = 9. Therefore aₙ = 5 + (n − 1)9 = 5 + 9n − 9 = 9n − 4, so option A is correct. Substitution provides several checks: for n = 1, the rule gives 9 − 4 = 5; for n = 2, it gives 18 − 4 = 14; and for n = 3, it gives 27 − 4 = 23. Option B has the correct coefficient of n but an incorrect constant, while options C and D do not reproduce both the first term and the common difference. Thus only A represents the given progression for every positive integer n.

The governing condition for a positive term is aₙ > 0. Applying the explicit rule gives 5n − 12 > 0, so 5n > 12 and n > 12/5 = 2.4. Since n must be a positive integer, the smallest possible index satisfying this inequality is n = 3. Direct evaluation confirms the transition: a₁ = 5 − 12 = −7, a₂ = 10 − 12 = −2, and a₃ = 15 − 12 = 3. Thus the third term is the first positive term, so option B is correct. Although the fourth and fifth terms are also positive, they occur later and therefore cannot be the first. Option A is incorrect because the second term remains negative. The inequality and the direct check agree.

The governing concept is the nth-term formula for an arithmetic progression, aₙ = a₁ + (n − 1)d. Here a₁ = 13 and the common difference is d = 18 − 13 = 5. For n = 20, a₂₀ = 13 + (20 − 1)5 = 13 + 95 = 108. Therefore option B is correct. The indexing is important: from the first term to the twentieth term there are 19 intervals, not 20, because the first term is already counted as the starting value. Using 18 intervals produces 103, while using 20 intervals produces 113. The last listed terms also show the pattern increasing by 5, and repeated addition for 19 steps reaches 108. Thus the formula and the sequence pattern agree.

The governing concept is the nth-term formula for an arithmetic progression: a_n = a + (n - 1)d. The first term is a = 13, and the common difference is d = 19 - 13 = 6. For the fourteenth term, use n = 14: a_14 = 13 + (14 - 1)6 = 13 + 78 = 91. Hence option B is correct. The number of intervals from the first term to the fourteenth term is 13, which explains the factor (14 - 1). Option A results from using only 12 intervals, and option C results from using 14 intervals instead of 13. Option D uses an incorrect difference or index. Substitution into the standard formula gives an exact and unambiguous result.

The governing concept is the explicit or general rule of an arithmetic progression. The first term is a = 9 and the common difference is d = 15 − 9 = 6. Hence aₙ = a + (n − 1)d = 9 + 6n − 6 = 6n + 3. Substituting n = 1 gives 9, n = 2 gives 15, and n = 3 gives 21, so the rule reproduces the sequence. Therefore, option A is correct. Option B gives 9 for the first term but then increases by 9, not 6. Option C gives 3 as its first term, and option D gives 9 initially but has common difference 3. Checking both the first term and the difference removes these distractors.

This is an nth-term problem. The first term is a = 64 and the common difference is d = 57 − 64 = −7. Use aₙ = a + (n − 1)d and set the term equal to 8: 8 = 64 + (n − 1)(−7). Rearranging gives 7(n − 1) = 56, so n − 1 = 8 and n = 9. Therefore 8 is the ninth term, making option B correct. The negative common difference means that the progression decreases by 7 at each step; it does not alter the nth-term formula. A direct check is 64 − 8 × 7 = 64 − 56 = 8, confirming the ninth position. Options A, C, and D would correspond to using too few or too many decreases.

The governing concept is the nth-term formula for an arithmetic progression, aₙ = a + (n − 1)d. The first term is a = 17, and the common difference is d = 22 − 17 = 5. For n = 25, there are 25 − 1 = 24 differences after the first term. Thus a₂₅ = 17 + 24(5) = 17 + 120 = 137, so option B is correct. The number 24 is essential: the first term has zero jumps, while the 25th term has twenty-four jumps of size 5. Option A would result from using one fewer difference, option C from adding one extra difference, and option D from a larger calculation error. Substitution of n = 1 gives 17, confirming the formula is indexed correctly.

The governing concept is the general-term formula for an arithmetic progression: aₙ = a₁ + (n − 1)d. The first term is a₁ = 12 and the common difference is d = 20 − 12 = 8. Hence aₙ = 12 + (n − 1)8 = 12 + 8n − 8 = 8n + 4. Therefore option A is correct. Checking the formula gives 12 when n = 1, 20 when n = 2, 28 when n = 3, and 36 when n = 4. Option C gives 4 as the first term, option B has common difference 12, and option D gives 12 for n = 1 but only 16 for n = 2. Thus only option A generates every term in the stated progression.

Direct answer: option B, 22. The first term is a₁ = x and the common difference is d = 9. Using aₙ = a₁ + (n − 1)d, the fifth term is a₅ = x + (5 − 1)9 = x + 36. Since a₅ = 58, solve x + 36 = 58, giving x = 22. Writing the terms explicitly confirms this: x, x + 9, x + 18, x + 27, x + 36. With x = 22, the fifth term is 58. If x = 20, the fifth term is 56; if x = 24, it is 60; and if x = 26, it is 62. Thus A, C, and D fail. The common mistake is to add 9 only three times; reaching the fifth term requires four jumps.

Direct answer: 0 is the 10th term, so option C is correct. An arithmetic progression changes by the same amount each time. Here the common difference is d = 64 − 72 = −8, meaning 8 is subtracted at every step. The nth-term formula is aₙ = a + (n − 1)d. Substituting a = 72 and d = −8 gives aₙ = 72 − 8(n − 1). Set this equal to zero: 72 − 8(n − 1) = 0, so 8(n − 1) = 72, n − 1 = 9, and n = 10. Option A is wrong because the 8th term is 16. Option B is wrong because the 9th term is 8. Option D is wrong because the 11th term is −8. A useful check is to count: 72, 64, 56, 48, 40, 32, 24, 16, 8, 0. Memory cue: when the difference is negative, continue subtracting; do not count the first term as a subtraction step.

The governing condition for a positive term is a_n > 0. Using the given rule, 7n - 23 > 0, so 7n > 23 and n > 23/7, which is approximately 3.29. The smallest positive integer position satisfying this is n = 4. Direct checking is also useful: a_3 = 7(3) - 23 = -2, which is not positive, whereas a_4 = 7(4) - 23 = 5, which is positive. Since the sequence increases by 7 each time, every later term is also positive, so the first positive term is the fourth term. Therefore option B is correct; options C and D identify later positive terms, not the first one.

Use the arithmetic-progression relation a_n = a + (n - 1)d. The first term is a = 18 and the common difference is d = 25 - 18 = 7. Hence a_n = 18 + 7(n - 1) = 7n + 11. Since the question states that a_n = 102, solve 7n + 11 = 102. Subtracting 11 from both sides gives 7n = 91, and division by 7 gives n = 13. Thus 102 is the thirteenth term, so option C is correct. A direct verification is 18 + (13 - 1)7 = 18 + 84 = 102. The nearby choices 11, 12, and 14 do not satisfy the equation; they result from an arithmetic error or from counting the number of differences incorrectly.

The governing relation is a_n = a_1 + (n - 1)d. For n = 7, the seventh term is six common differences after the first term, so a_7 = a_1 + 6d. Substitute the known values: 50 = a_1 + 6(9) = a_1 + 54. Solving gives a_1 = 50 - 54 = -4, so option A is correct. A backward check gives the same answer: starting at 50 and subtracting 9 six times produces 41, 32, 23, 14, 5, and finally -4. Options B, C, and D fail because they do not allow six increases of 9 to reach 50 at the seventh position. The key point is that the first and seventh terms are separated by 7 - 1 = 6 differences, not seven.

For an arithmetic progression, use a_n = a + (n - 1)d. The first term is a = 19 and the common difference is d = 25 - 19 = 6. For n = 24, a_24 = 19 + (24 - 1)6 = 19 + 23(6). Since 23 × 6 = 138, the value is 19 + 138 = 157. Therefore option B is correct. The use of 23 differences is important: the first term is already 19, so only 24 - 1 steps are needed to reach the twenty-fourth term. Option A is obtained by adding too few differences, while C and D result from adding one or more extra differences or making an arithmetic error. Substitution into the formula gives an exact verification.

The governing concept is the nth-term formula for a geometric progression: aₙ = a₁rⁿ⁻¹. Here a₁ = 12, r = 1/2, and n = 4, so a₄ = 12(1/2)³ = 12 × 1/8 = 12/8 = 3/2. Therefore, option A is correct. The same result can be seen by generating the terms one at a time: a₁ = 12, a₂ = 6, a₃ = 3, and a₄ = 3/2. Option D, 6, is only the second term; option C, 3, is the third term; and option B, 2, does not follow from three successive multiplications by 1/2. The exponent n − 1 is essential because the first term requires zero applications of the ratio.

The governing formula is aₙ = a₁rⁿ⁻¹. For the fifth term, the exponent is 5 − 1 = 4, so 162 = 2r⁴. Dividing both sides by 2 gives r⁴ = 81 = 3⁴. The real fourth-root solutions are r = 3 and r = −3, but the question explicitly states that the ratio is positive. Therefore r = 3, making option B correct. A common error is to use exponent 5 instead of 4, even though moving from the first term to the fifth requires only four multiplications. Substituting the other positive options gives 2(2⁴) = 32, 2(4⁴) = 512, and 2(5⁴) = 1250, none of which equals 162.

The consecutive differences are 9 − 4 = 5, 14 − 9 = 5, and 19 − 14 = 5, so the sequence is arithmetic with first term a₁ = 4 and common difference d = 5. The governing formula is aₙ = a₁ + (n − 1)d. Substituting the values gives aₙ = 4 + (n − 1)5 = 4 + 5n − 5 = 5n − 1. Therefore option A is correct. Verification is important: n = 1 gives 4, n = 2 gives 9, and n = 4 gives 19. Option B gives 6 as its first term; option C begins with 9 and has the wrong difference; option D has a difference of only 1. Thus only option A reproduces both the starting value and the constant increase.

To find the eighth term, substitute \(n=8\) in \(a_n=3n+2\): \(a_8=3\times8+2=24+2=26\). Hence, \(26\) is correct. \(24\) is only \(3\times8\); the constant term \(+2\) must also be added. Exam tip: After substituting the term number for \(n\), include every constant term and sign carefully.

Here \(a=18\) and \(d=-4\). Using \(a_n=a+(n-1)d\) gives \(18-4(n-1)\). The expression \(18-4n\) gives 14 when \(n=1\), not the first term 18. Exam tip: always test a proposed nth-term rule with \(n=1\).

In this geometric progression, the first term is \(a=2\) and the common ratio is \(r=6/2=3\). The \(n\)th term is \(a_n=ar^{n-1}\). Therefore, \(a_6=2\times 3^{6-1}=2\times 3^5=486\). Although 243 equals \(3^5\), it misses multiplication by the first term, 2. Exam tip: In a GP, the exponent is \(n-1\), not \(n\).