Word problems based on APs

Correct answer: C. (140)

Using (S_{15}=3150) and (d=10) gives (a=140). Exam tip: the first term can be found from total sum and difference.

Correct answer: B. (455)

The seats are (13,16,19,\ldots) and (S_{14}=455). Exam tip: convert real-life seating questions into APs.

Correct answer: C. (2550) रुपये

The savings form the AP (120,150,180,\ldots) and (S_{10}=2550). Exam tip: use the sum formula for total amount.

Correct answer: D. (90)

The seats form an AP and (a_{18}=22+17(4)=90). Exam tip: treat the row number as the term number.

Correct answer: C. (2000)

The production is the AP (180,200,220,\ldots) and (S_8=2000). Exam tip: treat the daily increase as the common difference.

Correct answer: B. (15)

Putting (S_n=360) gives (n=15). Exam tip: when total sum is given, form an equation for (n).

Correct answer: A. (675)

The production is the decreasing AP (95,90,85,\ldots) and (S_9=675). Exam tip: treat a decrease as a negative common difference.

Correct answer: B. (50)

Using (S_{10}=725) and (d=5) gives (a=50). Exam tip: solve the sum formula for the unknown first term.

Correct answer: C. (11)

From (18+(n-1)7=88), (n=11). Exam tip: set (a_n) equal to the given number to find the required day.

Correct answer: C. (632)

The seats are (17,20,23,\ldots) and (S_{16}=632). Exam tip: you may also find the last row and use (S_n=\frac{n}{2}(a+l)).

Correct answer: C. (280) लीटर

The direct answer is C, 280 litres. This is a decreasing arithmetic progression. The first day gives \\(a=640\\), each following day has 30 litres less, so \\(d=-30\\), and the requested day is \\(n=13\\). From day 1 to day 13 there are 12 decreases. Therefore \\(a_{13}=a+(n-1)d=640+12(-30)=640-360=280\\) litres. Option C is correct. Option A, 250 litres, subtracts too much; option B, 270 litres, does not follow the required 12 reductions; option D, 300 litres, subtracts too little. The common error is to multiply 30 by 13 instead of by 12, because the first day already starts at 640 and has not yet lost 30 litres. Memory cue: on the first day use the given amount; before the nth day, exactly \\(n-1\\) reductions occur.

Correct answer: B. 300 rupees

The governing concept is the sum of the first n terms of an arithmetic progression. The monthly deposits form an AP with first term a = 1,500, number of terms n = 7, total S7 = 16,800, and common difference d equal to the monthly increase. Use S_n = n/2[2a + (n − 1)d]. Substitution gives 16,800 = 7/2[3,000 + 6d]. Multiplying by 2 and dividing by 7 gives 4,800 = 3,000 + 6d, so 6d = 1,800 and d = 300. Hence option B is correct. The other choices result from an incorrect sum or an incorrect division; the required quantity is the common difference, not the final deposit.

Correct answer: C. (255) मीटर

The distances are (12,15,18,\ldots) and (S_{10}=255). Exam tip: the number of gaps and the number of poles are different.

Correct answer: B. (24000) रुपये

The salary forms an AP and (a_9=18000+8(750)=24000). Exam tip: find the monthly salary using the (n)th term.

Correct answer: A. (252)

The marks form the AP (5,7,9,\ldots) and (S_{14}=252). Exam tip: take the number of questions as (n) and find the sum.

Correct answer: C. (336)

The participant numbers are (21,27,33,\ldots) and (S_8=336). Exam tip: treat each round as one term.

Correct answer: C. (175)

The harvest is the AP (4,7,10,\ldots) and (S_{10}=175). Exam tip: use the sum formula for total area.

Correct answer: B. (8)

The amounts are (40,60,80,\ldots) and (S_n=1120) gives (n=8). Exam tip: when a total target is given, find the number of terms.

Correct answer: B. (20)

The direct answer is option B, 20 bulbs. The rows form an arithmetic progression because the same number is added or subtracted each time. The first row has 60 bulbs, and moving upward reduces the number by 5, so the common difference is \(d=-5\). The ninth row is \(a_9=a+(9-1)d=60+8(-5)=60-40=20\). Therefore B is correct. Option A, 15, would result from subtracting 5 nine times instead of only eight times after the first row. Option B, 20, matches the calculation. Option C, 25, is too large because it subtracts only 35. Option D, 30, is also too large because it subtracts only 30. Remember: the first row is already counted, so the ninth term has eight changes.

Correct answer: B. (900)

The direct answer is option B, 900 rupees. Each kilometre has its own fare, so the fares are an arithmetic progression: 60, 70, 80, and so on. Here \(a=60\), \(d=10\), and \(n=9\). The ninth fare is \(a_9=60+8(10)=140\). The total is \(S_9=\frac{9}{2}(60+140)=\frac{9}{2}\times200=900\) rupees. Option A, 860, is not the correct sum of these nine fares. Option B, 900, follows directly from the AP sum formula. Option C, 940, is too high and does not match the calculated total. Option D, 980, is also too high. Do not multiply the ninth fare by 9; fares increase after every kilometre, so all nine different fares must be added.

Correct answer: A. (722)

The seats are (11,14,17,\ldots) and (S_{19}=722). Exam tip: use the usual sum formula even for an odd number of rows.

Correct answer: B. (500)

Using (S_5=17500) and (a=2500) gives (d=500). Exam tip: equal increase in instalments is the common difference.

Correct answer: C. (192)

The lane numbers are (5,7,9,\ldots) and (S_{12}=192). Exam tip: a sequence with a small constant increase is also an AP.

Correct answer: B. (180000)

The profits are (25000,27000,29000,\ldots) and (S_6=180000). Exam tip: view yearly profits as AP terms.

Correct answer: B. (11)

Putting (S_n=525) gives (n=11). Exam tip: when a total target is given, find the number of terms.