Word problems based on APs

The numbers of plants form an arithmetic progression with first term \(a=21\) and common difference \(d=11\). The number of plants in the \(n\)th row is \(a_n=21+(n-1)\times 11\). Setting this equal to 252 gives \(21+(n-1)\times 11=252\), so \(11(n-1)=231\), \(n-1=21\), and \(n=22\). Hence, the 22nd row has 252 plants. The 21st row would have only 241 plants. Exam tip: when a target term is given, equate it to \(a_n\) and solve for \(n\).

The daily work forms the AP (68,82,96,\ldots) and (S_{17}=3060). Exam tip: treat the daily increase as the common difference.

The amounts are (350,500,650,\ldots) and (S_n=12600) gives (n=12). Exam tip: set the target total equal to (S_n).

The numbers of stones form a decreasing AP from the bottom, with \(a=210\) and \(d=-10\). For the \(n\)th row, \(a_n=210+(n-1)(-10)\). Using \(210-10(n-1)=70\) gives \(n-1=14\), so \(n=15\). Hence, the 15th row has 70 stones. The 14th row would have 80 stones, so it is not correct. Exam tip: use a negative common difference for an AP that decreases upward.

The fares form the AP (130,170,210,\ldots) and (S_{10}=3100). Exam tip: treat each kilometre's fare as one term.

The matchstick numbers are (18,31,44,\ldots) and (a_{18}=239). Exam tip: take the figure number as the term number.

The deposits are (500,570,640,\ldots) and (S_{18}=19710). Exam tip: treat weekly deposits as AP terms.

The amounts form the decreasing AP (2400,2310,2220,\ldots) and (S_{16}=27600). Exam tip: the common difference is negative in a decreasing quantity.

The applications form the AP (35,47,59,\ldots) and (a_{16}=215). Exam tip: find the (n)th term for a particular month.

The capacities form the AP (54,63,72,\ldots) and (S_{20}=2790). Exam tip: add the capacities of all rows.

The pot numbers are (32,43,54,\ldots) and (S_{19}=2470). Exam tip: use the sum formula for total production.

The admissions are (72,90,108,\ldots) and (S_n=3780) gives (n=15). Exam tip: connect the target total with (S_n).

The production figures form an arithmetic progression. The first-hour production is the first term, \\(a=140\\), and the increase each hour is the common difference, \\(d=22\\). The 14th hour is the 14th term, so use \\(a_n=a+(n-1)d\\). Thus \\(a_{14}=140+(14-1)22=140+286=426\\) units.

Option C is therefore correct. There are 13 increases from the first hour to the fourteenth hour, because the first hour already starts at 140 units. Adding 22 fourteen times would incorrectly count one extra increase and produce 448. The sequence begins 140, 162, 184, and continues by adding 22 each time, which confirms the arithmetic. The supplied answer is correct and the instruction to treat the hour number as the term number is appropriate.

The work hours are (28,37,46,\ldots) and (S_{15}=1365). Exam tip: add all days using the sum formula.

The instalments are (9500,11000,12500,\ldots) and (S_7=98000). Exam tip: treat instalments as AP terms.

Using (S_{24}=2520) and (d=5) gives (a=47). Exam tip: find the first term from total and difference.

The sales form the decreasing AP (360,342,324,\ldots) and (S_{15}=3510). Exam tip: keep (d) negative for decreasing sales.

This forms an arithmetic progression with first term \(a=70\) and common difference \(d=28\). The number of pages read in the \(n\)th month is \(a_n=70+(n-1)28\). Setting \(70+(n-1)28=434\) gives \((n-1)28=364\), so \(n-1=13\) and \(n=14\). Therefore, the student will read 434 pages in the 14th month. The nearby option 13 is incorrect because the 13th-month reading is \(406\) pages. Exam tip: equate the target value to \(a_n\) and solve for \(n\).

The distances are (40,50,60,\ldots) and (S_{16}=1840). Exam tip: take the number of gaps as the number of terms.

The member numbers are (50,62,74,\ldots) and (S_{18}=2736). Exam tip: treat monthly new members as AP terms.