Word problems based on APs

Correct answer: C. (5)

Using (S_{10}=725) and (a=50) gives (d=5). Exam tip: total work and first day work can give the increase.

Correct answer: A. (11)

From (12+(n-1)6=72), (n=11). Exam tip: set (a_n) equal to the given number to find the required day.

Correct answer: C. (16)

The direct answer is C, 16 rows. The row sizes form an arithmetic progression: first term \\(a=30\\), common difference \\(d=5\\), and total \\(S_n=1080\\). Use \\(S_n=\frac n2[2a+(n-1)d]\\): \\(1080=\frac n2[60+5(n-1)] =\frac n2(5n+55)\\). Thus \\(2160=5n^2+55n\\), or \\(n^2+11n-432=0\\). Factoring gives \\((n-16)(n+27)=0\\), so \\(n=16\\) or \\(n=-27\\). A number of rows cannot be negative, hence 16. Option C is correct. Option A, 14, gives a smaller total; option B, 15, also does not reach 1080; option D, 18, gives a larger total. Always reject a negative mathematical root in a counting problem. Exam cue: after solving, substitute the positive integer back into the sum formula.

Correct answer: D. (10)

The amounts are (25,50,75,\ldots) and (S_n=1375) gives (n=10). Exam tip: treat multiple-based savings as a simple AP.

Correct answer: A. (140)

The charge increases are (7,10,13,\ldots) and (S_8=140). Exam tip: percentage increases can also be used as AP terms.

Correct answer: C. (70)

The tile numbers are (4,10,16,\ldots) and (a_{12}=70). Exam tip: take the figure number as the term number.

Correct answer: B. (10)

Using (S_5=300) and (a=40) gives (d=10). Exam tip: daily increase can be found from total distance.

Correct answer: A. (240)

The member numbers are (20,28,36,\ldots) and (S_6=240). Exam tip: treat monthly new members as AP terms.

Correct answer: C. (99)

The harvest is the AP (3,5,7,\ldots) and (S_9=99). Exam tip: use the sum formula for total area.

Correct answer: C. (820)

The direct answer is C: 820 rupees. Treat the fare charged for each kilometre as one term of an arithmetic progression. The first term is \\(a=50\\), the common difference is \\(d=15\\), and there are \\(n=8\\) terms. The eighth kilometre costs \\(a_8=50+7(15)=155\\). The total is the sum: \\(S_8=\\frac{8}{2}[2(50)+(8-1)(15)] =4(100+105)=4\\times205=820\\). Option A, 760, is too low and does not result from the correct eight-term sum. Option B, 790, is also an incorrect total, usually caused by using the wrong number of increases or an arithmetic error. Option C, 820, is correct because it adds the fares for kilometres 1 through 8. Option D, 850, is too high and may result from adding an extra increase. The important point is that the question asks for total fare, so finding only the eighth fare, 155, is not enough. Memory cue: first fare plus equal increase means AP; total means use the sum formula.

Correct answer: B. (119)

The lane numbers are (2,3,4,\ldots) and (S_{14}=119). Exam tip: a sequence with small constant increase is also an AP.

Correct answer: A. (100000)

The profits are (18000,19000,20000,\ldots) and (S_5=100000). Exam tip: view yearly profits as AP terms.

Correct answer: C. (12)

Putting (S_n=306) in the sum formula gives (n=12). Exam tip: you can also check options in the sum formula.

Correct answer: C. (775)

The production is the decreasing AP (100,95,90,\ldots) and (S_{10}=775). Exam tip: for decreasing production rate, (d) is negative.

Correct answer: D. (110)

The page numbers are (20,35,50,\ldots) and (a_7=110). Exam tip: find the (n)th term for a particular month.

Correct answer: B. (380)

The direct answer is B: 380 lights. The numbers of lights form an arithmetic progression: 30, 35, 40, 45, 50, 55, 60 and 65. The first term is a=30, the common difference is d=5, and the number of terms is n=8. The sum formula is S_n=\frac{n}{2}[2a+(n-1)d]. Therefore S_8=\frac{8}{2}[2(30)+7(5)]=4(60+35)=4\times95=380. Option A, 360, is too small and does not equal the sum of the eight terms. Option B, 380, matches the calculation and is correct. Option C, 400, is not the sum produced by the progression. Option D, 420, is also too large. Another quick check is pairing the terms: (30+65), (35+60), (40+55), (45+50), giving four pairs of 95, or 380. Treat each gate as one term; do not multiply the eighth term by eight.

Correct answer: B. (600)

Using (S_6=21000) and (a=2000) gives (d=600). Exam tip: equal increase in instalments is the common difference.

Correct answer: A. (260)

The work hours are (8,12,16,\ldots) and (S_{10}=260). Exam tip: add all days using the sum formula for total hours.

Correct answer: C. (348)

The packet numbers are (18,20,22,\ldots) and (S_{12}=348). Exam tip: take the number of rows as the number of terms.

Correct answer: B. (210)

The sales form the decreasing AP (60,50,40,\ldots) and (S_6=210). Exam tip: take (d) negative for decreasing sales.

Correct answer: D. (103)

The numbers form the AP (5,12,19,\ldots) and (a_{15}=103). Exam tip: work done on the (n)th day is the (n)th term.

Correct answer: A. (567)

The seats are (7,9,11,\ldots) and (S_{21}=567). Exam tip: use the usual sum formula even for an odd number of rows.

Correct answer: C. (12)

Putting (S_n=450) gives (n=12). Exam tip: when a total target is given, find the number of terms.

Correct answer: C. (195)

The deposits are (60,75,90,\ldots) and (a_{10}=195). Exam tip: use (a_n) for the amount on a particular day.

Correct answer: A. (550)

The box numbers are (25,30,35,\ldots) and (S_{11}=550). Exam tip: total can also be found using the average of first and last shelves.