Word problems based on APs

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.

Using (S_{20}=1740) and (d=6) gives (a=30). Exam tip: find the first term from total and difference.

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.

The video numbers are (10,16,22,\ldots) and (S_{24}=1896). Exam tip: take the number of weeks as the number of terms.

The lane numbers are (12,17,22,\ldots) and (S_{21}=1302). Exam tip: treat a sequence with equal increase as an AP.