Word problems based on APs

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

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

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.

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

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

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

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

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

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

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.

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

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

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

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

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

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.

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

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

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

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

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.

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.

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

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

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.