Word problems based on APs

The member numbers are (12,18,24,\ldots) and (S_{17}=1020). Exam tip: add all days for total members.

Using (S_{15}=720) and (a=20) gives (d=4). Exam tip: find the common difference from total and first term.

The plate numbers are (15,20,25,\ldots) and (S_{14}=665). Exam tip: treat each table as one term of the AP.

The doses form the AP (8,11,14,\ldots) and (S_{20}=730). Exam tip: add the terms for the total dose.

The water amounts are (100,125,150,\ldots) and (S_7=1225). Exam tip: treat hourly filled water as AP terms.

Here, the first term is 90 and the common difference is 10. Therefore, the number of newspapers sold on the nth day is
\(a_n=90+(n-1)\times10\). Putting
\(90+(n-1)\times10=230\) gives
\(n-1=14\), so
\(n=15\). Day 14 is a close distractor, but only 220 newspapers would be sold that day. Exam tip: equate the target value to the nth-term formula of the AP and solve for n.

Putting (S_n=1859) in the sum formula gives (n=22). Exam tip: when total seats are given, form an equation for (n).

The amounts form the AP (90,120,150,\ldots) and (S_{14}=3990). Exam tip: use the sum formula for total saving.

The production is the decreasing AP (360,345,330,\ldots) and (S_{16}=3960). Exam tip: treat a decrease as a negative common difference.

Using (S_{18}=972) and (a=20) gives (d=4). Exam tip: find the common difference from the total and first term.

The numbers of plants form an arithmetic progression with first term 25 and common difference 8. Therefore, the number of plants in the nth row is \(a_n=25+(n-1)\times8\). Putting \(a_n=153\), we get \(25+(n-1)\times8=153\), so \(8(n-1)=128\) and hence \(n=17\). The 18th row would contain \(161\) plants, so it is not correct. Exam tip: equate the given target value to \(a_n\) and solve for \(n\).

The daily work forms the AP (55,64,73,\ldots) and (S_{18}=2367). Exam tip: treat the daily increase as the common difference.

The donation amounts are (200,300,400,\ldots) and (S_n=5400) gives (n=9). Exam tip: form an (S_n) equation for the target total.

The number of tiles forms a decreasing AP from the bottom row, with first term \(a=132\) and common difference \(d=-6\). Thus, the number of tiles in the \(n\)th row is \(a_n=132+(n-1)(-6)\). Using \(132-6(n-1)=54\), we get \(6(n-1)=78\), so \(n-1=13\) and \(n=14\). Therefore, the 14th row has 54 tiles. The 13th row has 60 tiles, so it is a close but incorrect option. Exam tip: use a negative common difference for a decreasing AP.

The fares are (100,125,150,\ldots) and (S_9=1800). Exam tip: treat the fare for each kilometre as one term.

The matchstick numbers are (15,24,33,\ldots) and (a_{20}=186). Exam tip: take the figure number as the term number.

The deposits are (300,345,390,\ldots) and (S_{16}=10200). Exam tip: treat weekly deposits as AP terms.

The amounts form the decreasing AP (1500,1440,1380,\ldots) and (S_{12}=14040). Exam tip: keep (d) negative for decreasing quantities.

The salary forms an AP and (a_{12}=46700). Exam tip: find a particular month's salary using the (n)th term.

The capacities are (42,48,54,\ldots) and (S_{20}=1980). Exam tip: add all rows for total capacity.