Word problems based on APs

This forms an arithmetic progression with first term 12 and common difference 5. Hence, \(a_n=12+(n-1)\times5\). Putting \(a_n=77\), we get \(12+5(n-1)=77\), so \(5(n-1)=65\), \(n-1=13\), and \(n=14\). Therefore, 77 instruments will be placed on the 14th day. On the 13th day, there would be only 72 instruments. Exam tip: in AP word problems, identify the first term and common difference, then equate the target value to \(a_n\).

The governing concept is the nth-term formula for an arithmetic progression. Hourly production forms the sequence 150, 140, 130, and so on, so the first term is a = 150 and the common difference is d = -10 because production decreases by 10 each hour. For the nth hour, a_n = a + (n - 1)d. Setting the production equal to 60 gives 60 = 150 + (n - 1)(-10). Thus 60 = 150 - 10n + 10 = 160 - 10n, so 10n = 100 and n = 10. Therefore option D is correct. Options A, B and C represent earlier hours and give 90, 80 and 70 packets respectively, not 60.

The correct answer is 28 seats, option C. Let the number of seats in the first row be a. The rows form an AP with n = 16, common difference d = 4, and total S_16 = 928. Apply S_n = n/2[2a + (n − 1)d]: 928 = 16/2[2a + 15 × 4] = 8(2a + 60). Dividing by 8 gives 116 = 2a + 60, so 2a = 56 and a = 28. Therefore the first row had 28 seats. Option A, 24, option B, 26, and option D, 30, would give totals different from 928. Check the result: the last row is 28 + 15 × 4 = 88 seats, and the average row size is (28 + 88)/2 = 58; 16 × 58 = 928. This confirms the answer and illustrates why the AP sum formula is suitable.

This is an arithmetic-progression word problem involving the nth term. The first-floor count is the first term, a = 150, and the number of windows decreases by 10 on every successive floor, so the common difference is d = −10. For the nth floor, use aₙ = a + (n − 1)d. Setting the required count equal to 60 gives 60 = 150 + (n − 1)(−10). Thus −90 = −10(n − 1), so n − 1 = 9 and n = 10. Therefore, option C is correct. The sequence is 150, 140, 130, 120, 110, 100, 90, 80, 70, 60. Option A would give 80 windows, option B 70, and option D 50, so none of them satisfies the condition.

The governing concept is the sum of the terms of an arithmetic progression. The daily incomes form 700, 850, 1000, and so on, so the first term is a = 700, the common difference is d = 150, and the number of terms is n = 8. Use Sₙ = n/2[2a + (n − 1)d]. Therefore, S₈ = 8/2[2(700) + 7(150)] = 4[1400 + 1050] = 4 × 2450 = 9800 rupees. Hence option B is correct. The question asks for the total of all eight days, not merely the income on day eight. The eighth-day income would be 700 + 7(150) = 1750 rupees, which is only one term. The other choices arise from incomplete or incorrect addition.

This problem uses the nth-term formula for a decreasing arithmetic progression. The first term is a = 180 because the first floor has 180 windows. Since every higher floor has 15 fewer windows, the common difference is d = −15. For the nth floor, aₙ = a + (n − 1)d. Put aₙ = 45: 45 = 180 + (n − 1)(−15). Hence −135 = −15(n − 1), so n − 1 = 9 and n = 10. Thus option C is correct. A direct check gives 180 on floor 1, 165 on floor 2, and after nine decreases, 180 − 9×15 = 45 on floor 10. Floor 8 has 75, floor 9 has 60, and floor 11 has 30 windows, so the distractors do not meet the requirement.

This is an arithmetic-progression word problem because the number of seats increases by the same amount, 7, from one row to the next. The first term is a = 34, the common difference is d = 7, and the number of rows is n = 21. Since the question asks for the total, use S_n = n/2[2a + (n − 1)d]. Substitution gives S_21 = 21/2[2(34) + 20(7)] = 21/2[68 + 140] = 21/2 × 208 = 21 × 104 = 2184. Therefore, option C is correct. The distractors can result from using the nth-term formula instead of the sum formula, omitting a row, or making an arithmetic error. The constant increase confirms that AP methods are appropriate.

The row sizes form an arithmetic progression because each row increases by the same amount. The first term is a = 25, the number of rows is n = 22, and the total is S_22 = 1243. Apply S_n = n/2[2a + (n − 1)d]: 1243 = 22/2[2(25) + 21d] = 11(50 + 21d). Dividing by 11 gives 113 = 50 + 21d. Hence 21d = 63 and d = 3. Therefore, option B is correct. This value can be checked directly: the last row is 25 + 21(3) = 88, and the average row size is (25 + 88)/2 = 56.5; multiplying by 22 gives 1243. The other options fail the given total.

The governing concept is the nth-term formula of an arithmetic progression. The first row gives a_1 = 33 and the 18th row gives a_18 = 152. For an AP, a_n = a + (n − 1)d, where d is the equal increase. Substituting the known values gives 152 = 33 + (18 − 1)d = 33 + 17d. Therefore, 119 = 17d and d = 7. Hence option C is correct. A quick check confirms it: starting at 33 and adding 7 seventeen times gives 33 + 119 = 152. If the increase were 5, 6, or 8, the 18th-row counts would be 118, 135, or 169, respectively, so those distractors cannot satisfy the data.

The successive gaps are 18, 27, 36, and so on, so they form an arithmetic progression. The first term is a = 18, the common difference is d = 9, and the number of gaps is n = 17. Because the question asks for total distance, not just the last gap, use S_n = n/2[2a + (n − 1)d]. Thus S_17 = 17/2[2(18) + 16(9)] = 17/2[36 + 144] = 17/2 × 180 = 17 × 90 = 1530 metres. Therefore, option A is correct. The 17th gap alone would be 18 + 16×9 = 162 metres, so confusing a_n with S_n could lead to an incorrect distractor. The other values also do not satisfy the required sum.

The seat counts form an arithmetic progression with first term a = 36 and common difference d = 6. If there are n rows, the total is S_n = n/2[2a + (n − 1)d]. Substituting the data gives 1566 = n/2[72 + 6(n − 1)] = n/2(6n + 66) = 3n(n + 11). Therefore, n(n + 11) = 522. Testing the positive whole-number choices, n = 18 gives 18(18 + 11) = 18 × 29 = 522, so the number of rows is 18. Option D is correct. Direct verification gives the last row as 36 + 17×6 = 138; the average of the first and last rows is 87, and 87×18 = 1566. The other choices do not produce the stated total.

The governing concept is the sum of the terms of an arithmetic progression. The daily numbers of instruments form an AP with first term a = 16, number of days n = 25, total S_n = 1600, and common difference d. Use S_n = n/2[2a + (n − 1)d]. Thus 1600 = 25/2[2(16) + 24d] = 25/2(32 + 24d). Multiplying by 2 and dividing by 25 gives 128 = 32 + 24d. Hence 24d = 96 and d = 4. Therefore, the number placed increased by four instruments each day, so option C is correct. The other choices would produce different totals when substituted into the same sum formula. This is a sum-based AP problem, not merely an nth-term question.

The numbers of windows form an arithmetic progression with first term a = 220 and common difference d = −20, because the number decreases by 20 on each successive floor. The nth-term formula is aₙ = a + (n − 1)d. Substituting the required number 40 gives 40 = 220 + (n − 1)(−20). Hence, −180 = −20(n − 1), so n − 1 = 9 and n = 10. Therefore, the tenth floor has 40 windows, making option C correct. Checking the sequence confirms this: floor 1 has 220 windows, floor 2 has 200, and after nine decreases of 20, floor 10 has 40. The other choices correspond to 60, 80 or 20 windows.

The governing concept is the nth term of an arithmetic progression with a negative common difference. The first floor gives a = 260, and because each successive floor has 18 fewer windows, d = −18. For the nth floor, a_n = a + (n − 1)d. Setting the term equal to 62 gives 62 = 260 − 18(n − 1). Hence −198 = −18(n − 1), so n − 1 = 11 and n = 12. A direct check confirms this: from the first to the twelfth floor there are 11 decreases, and 260 − 11 × 18 = 260 − 198 = 62. Thus option C is correct. The other answers result from counting 12 decreases instead of 11 or mishandling the negative difference.