Word problems based on APs

The direct answer is option C, row 15. This is a decreasing arithmetic progression because every upper row has 8 fewer tiles. The first row has 180 tiles, so \(a=180\), \(d=-8\), and the nth row is \(a_n=a+(n-1)d\). Put the required number 68 in the formula: \(68=180+(n-1)(-8)\). Therefore \(-112=-8(n-1)\), so \(14=n-1\), and \(n=15\). Option A, row 13, has \(180-12\times8=84\) tiles, not 68. Option B, row 14, has \(180-13\times8=76\) tiles. Option C, row 15, has \(180-14\times8=68\), so it is correct. Option D, row 16, has \(180-15\times8=60\) tiles. The negative common difference is important because the number decreases upward. Memory cue: in a decreasing AP, write the difference with a minus sign before solving.

The fares are (120,155,190,\ldots) and (S_{12}=3750). Exam tip: treat each kilometre's fare as one term.

The matchstick numbers are (22,33,44,\ldots) and (a_{19}=220). Exam tip: take the figure number as the term number.

The deposits are (450,510,570,\ldots) and (S_{15}=13050). Exam tip: treat weekly deposits as AP terms.

The amounts form the decreasing AP (1800,1725,1650,\ldots) and (S_{14}=18375). Exam tip: keep (d) negative for decreasing quantities.

The salary forms an AP and (a_{11}=53000). Exam tip: find a particular month's salary using the (n)th term.

The capacities are (48,56,64,\ldots) and (S_{18}=2088). Exam tip: add all rows for total capacity.

The pot numbers are (24,33,42,\ldots) and (S_{18}=1809). Exam tip: use the sum formula for total production.

The admissions are (60,80,100,\ldots) and (S_n=3000) gives (n=15). Exam tip: connect the target total with the sum formula.

The first-hour production is the first term of an arithmetic progression. Here the first term is \\(a=120\\), the common difference is \\(d=18\\), and the 16th hour means the 16th term. The nth-term formula is \\(a_n=a+(n-1)d\\). Substituting the values gives \\(a_{16}=120+(16-1)18=120+270=390\\) units.

Therefore option C is correct. The phrase “each next hour” tells us that 18 units are added repeatedly, and the first hour must not be counted as an added step. There are 15 increases between the first and sixteenth hours, not 16. This is why simply calculating \\(120+16(18)\\) would give an incorrect value of 408. The supplied answer and explanation correctly identify the sequence and its 16th term.

The work hours are (25,32,39,\ldots) and (S_{13}=871). Exam tip: add all days using the sum formula for total hours.

The instalments are (8000,9200,10400,\ldots) and (S_8=97600). Exam tip: treat instalments as AP terms.

The direct answer is B: 30 seats. Let the first row have a seats. There are 20 rows and each next row has 6 more seats, so the last row has a+19(6). The sum of an AP is S_n=n/2[2a+(n-1)d]. Substituting S_20=1740 and d=6 gives 1740=20/2[2a+19(6)]=10(2a+114). Thus 174=2a+114, so 2a=60 and a=30. Option B is correct. Option A, 28, would give total 1700; option C, 32, would give 1780; option D, 34, would give 1820. The common mistake is to divide the total by 20 and call that the first row; 87 is the average, not the first term. Since the rows increase, the first row must be below the average.

The sales form the decreasing AP (300,284,268,\ldots) and (S_{13}=2652). Exam tip: keep (d) negative for decreasing sales.

This forms an arithmetic progression with first term \(a=55\) and common difference \(d=25\). Pages read in the \(n\)th month are \(a_n=55+(n-1)25\). Setting this equal to 355 gives \(55+(n-1)25=355\), so \(25(n-1)=300\), \(n-1=12\), and \(n=13\). Hence, the student reads 355 pages in the 13th month. In the 12th month, the student would read only \(330\) pages. Exam tip: use \(a_n=a+(n-1)d\) when the quantity for a particular term is given.

The distances are (32,40,48,\ldots) and (S_{17}=1632). Exam tip: take the number of gaps as the number of terms.

The member numbers are (42,51,60,\ldots) and (S_{14}=1407). Exam tip: treat monthly new members as AP terms.

The passenger numbers are (15,23,31,\ldots) and (S_{18}=1494). Exam tip: take the number of stops as the number of terms.

The profits form the AP (90000,102000,114000,\ldots) and (S_8=1056000). Exam tip: treat yearly profits as separate terms.

The tile numbers are (18,27,36,\ldots) and (S_{16}=1368). Exam tip: find the total of a pattern using the sum formula.

The sales form the AP (180,205,230,\ldots) and (S_{11}=3355). Exam tip: treat daily sales as AP terms.

Using (S_{18}=15075) and (a=200) gives (d=75). Exam tip: find the common difference from total and first term.

The amount filled each minute forms an AP with \(a=44\) and \(d=13\). Thus, the amount filled in the \(n\)th minute is \(a_n=44+(n-1)13\). Solving \(44+(n-1)13=239\) gives \((n-1)13=195\), so \(n-1=15\) and \(n=16\). Therefore, 239 litres are filled in the 16th minute. In the 15th minute, the amount is \(226\) litres, making it a close but incorrect option. Exam tip: for “in which minute,” use \(a_n\), not the cumulative sum \(S_n\).

The patient numbers are (55,64,73,\ldots) and (S_{20}=2810). Exam tip: use the sum formula for total patients.

The sack numbers are (72,84,96,\ldots) and (S_{15}=2340). Exam tip: treat each row as one term of the AP.