Finding the $n$th term of an AP

The term a₁₃ is midway between a₇ and a₁₉ because the indices are equally spaced: 13 − 7 = 19 − 13 = 6. In an AP, the middle term is the average of the two equally spaced terms, so 2a₁₃ = 156 and a₁₃ = 78.

The first term of the AP is 128 and the common difference is \(d=-9\). The term after 38 is \(38-9=29\). Since \(38>30\) but \(29<30\), 38 is the last term greater than 30. Option 29 is a close distractor, but it is not greater than 30. Exam tip: In a decreasing AP, check the next term to confirm that it crosses the given limit.

A key property of an arithmetic progression is that terms with equally spaced indices have the middle term equal to the arithmetic mean of the outer terms. Here 18 − 9 = 9 and 27 − 18 = 9, so the 18th term is exactly midway between the 9th and 27th terms. Therefore a₁₈ = (a₉ + a₂₇)/2 = (46 + 154)/2 = 200/2 = 100. Hence option B is correct. The result can also be verified by finding the common difference: a₂₇ − a₉ = 18d, so 154 − 46 = 108 and d = 6. Moving nine steps from a₉ gives a₁₈ = 46 + 9(6) = 100. The other options do not satisfy the midpoint relation and cannot be the required term.

In an arithmetic progression, the term whose index is halfway between two equally spaced indices is the arithmetic mean of the corresponding terms. Since 22 − 10 = 12 and 34 − 22 = 12, the 22nd term lies exactly midway between the 10th and 34th terms. Hence a₂₂ = (a₁₀ + a₃₄)/2 = (54 + 174)/2 = 228/2 = 114. Therefore option B is correct. Algebraically, if aₙ = a + (n − 1)d, then a₁₀ = a + 9d and a₃₄ = a + 33d. Their average is a + 21d, which is exactly a₂₂. This confirms the midpoint method. Options A, C, and D do not equal the required average and therefore cannot be correct.

The daily practice times form an arithmetic progression: 36, 43, 50, and so on. Its first term is a₁ = 36 and its common difference is d = 7. Since the question asks for the time on one particular day, we use the nth-term formula rather than a sum formula: aₙ = a₁ + (n − 1)d. Set the term equal to 218: 218 = 36 + 7(n − 1). Thus 182 = 7(n − 1), so n − 1 = 26 and n = 27. Therefore the practice time is 218 minutes on the 27th day, so option C is correct. The nearby alternatives result from an indexing mistake, such as counting the first day as day zero, or from adding or subtracting one day.

In an arithmetic progression, terms whose indices are equally spaced have the middle term equal to the arithmetic mean of the two outer terms. The indices 12, 24, and 36 are equally spaced because 24 − 12 = 12 and 36 − 24 = 12. Hence 2a₂₄ = a₁₂ + a₃₆. Substituting the given values gives 2a₂₄ = 75 + 219 = 294, so a₂₄ = 294/2 = 147. Therefore option B is correct. This can also be checked using the common difference: 24 index steps separate a₁₂ and a₃₆, so d = (219 − 75)/(36 − 12) = 144/24 = 6; then a₂₄ = 75 + (24 − 12) × 6 = 147. The other options are not the required midpoint.

The governing concept is the nth-term form of an arithmetic progression. In a_n=kn−7, the coefficient k is the common difference because increasing n by 1 increases the term by k. Use the given difference: a_17−a_5=(17k−7)−(5k−7)=12k, since the two constant terms cancel. Therefore 12k=96 and k=8. Now evaluate the required term directly: a_29=29k−7=29×8−7=232−7=225. Hence option B is correct. A frequent mistake is to treat −7 as part of the common difference or to subtract it incorrectly; it cancels when two terms are subtracted. Options A, C, and D do not satisfy the expression with k=8, so they are numerical distractors rather than valid alternatives.

The sequence is an arithmetic progression with first term a₁ = 19 and common difference d = 30 − 19 = 11. Its nth term is aₙ = a₁ + (n − 1)d = 19 + 11(n − 1) = 11n + 8. We need the least index for which the term is strictly greater than 700, so 11n + 8 > 700. This gives 11n > 692 and n > 62.909..., hence the least integer index is n = 63. The corresponding term is a₆₃ = 19 + (63 − 1) × 11 = 19 + 682 = 701. The preceding term is a₆₂ = 19 + 61 × 11 = 690, which is not greater than 700. Thus option B is the first qualifying term; 712 and 723 occur later.

In an arithmetic progression, if one index is exactly midway between two other indices, its term is the arithmetic mean of the corresponding outer terms. Here 26 − 14 = 12 and 38 − 26 = 12, so a₂₆ is midway between a₁₄ and a₃₈. Thus 2a₂₆ = a₁₄ + a₃₈, giving a₂₆ = (92 + 284)/2 = 376/2 = 188. Option B is therefore correct. An independent check finds the common difference: d = (284 − 92)/(38 − 14) = 192/24 = 8. Moving from the fourteenth to the twenty-sixth term takes 12 steps, so a₂₆ = 92 + 12 × 8 = 188. The other choices do not satisfy this midpoint relation.

The first term is a₁ = 27 and the common difference is d = 40 − 27 = 13. Therefore the nth term is aₙ = 27 + (n − 1)13. We need the least n such that this term is strictly greater than 850: 27 + 13(n − 1) > 850. Hence 13(n − 1) > 823, so n − 1 > 63.307..., and the least integer value is n − 1 = 64, or n = 65. The term is a₆₅ = 27 + 64 × 13 = 27 + 832 = 859. The preceding term is a₆₄ = 27 + 63 × 13 = 846, which is not greater than 850. Thus option B is the first qualifying term. Options C and D are later terms, while A fails the strict inequality.

First calculate the common difference from consecutive terms. We have d = −15/2 − (−17) = −15/2 + 34/2 = 19/2. Checking the next pair gives 2 − (−15/2) = 4/2 + 15/2 = 19/2, so the progression is consistent. For an AP, aₙ = a₁ + (n − 1)d. Therefore a₃₅ = −17 + (35 − 1)(19/2) = −17 + 34 × 19/2 = −17 + 17 × 19 = −17 + 323 = 306. Hence option C is correct. The values 272, 276 and 284 do not satisfy the standard nth-term calculation. Including 306 as option C corrects the incomplete original option set and makes the answer unique.

The governing concept is the nth term of an arithmetic progression. The hourly production is 20, 25, 30, and so on, because the same difference of 5 packets is added each hour. Thus the first term is a = 20, the common difference is d = 5, and the required term number is n = 10. Applying aₙ = a + (n − 1)d gives a₁₀ = 20 + (10 − 1) × 5 = 20 + 45 = 65 packets. Therefore option C is correct. The tenth hour is reached after nine increases from the first hour, not ten. Options A and B count too few increases, while D counts too many. Since the question asks for production in one hour, a term formula, not a sum formula, is required.