Word problems based on APs

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.

The pot numbers are (18,25,32,\ldots) and (S_{16}=1128). Exam tip: use the sum formula for total production.

The admissions are (45,60,75,\ldots) and (S_n=2520) gives (n=16). Exam tip: connect the target total with the sum formula.

This is an arithmetic progression because the production increases by the same number, 15 units, every hour. The first term is \\(a=95\\), the common difference is \\(d=15\\), and the fifteenth hour represents the fifteenth term. The formula is \\(a_n=a+(n-1)d\\). Substituting the values gives \\(a_{15}=95+(15-1)\\times15=95+210=305\\). Thus the machine produces 305 units in the fifteenth hour.

Option B is correct. A common mistake is to multiply 15 by 15 without accounting for the first hour, or to use 15 increases instead of the 14 increases that occur before the fifteenth hour. The sequence begins 95, 110, 125, and continues by adding 15. Since reaching the fifteenth term requires fourteen additions, 305 is the required production.

The work hours are (20,25,30,\ldots) and (S_{14}=735). Exam tip: add all days using the sum formula for total hours.

The instalments are (6000,7000,8000,\ldots) and (S_7=63000). Exam tip: treat instalments as AP terms.

The correct answer is B. Let the first row have \\(a\\) seats. Here \\(n=18\\), total \\(S_{18}=1197\\), and common difference \\(d=5\\). Use \\(S_n=\frac{n}{2}[2a+(n-1)d]\\). Thus \\(1197=\frac{18}{2}[2a+17(5)] =9(2a+85)\\). Dividing by 9 gives \\(133=2a+85\\), so \\(2a=48\\) and \\(a=24\\). Hence option B is correct. Option A, 22, would give a smaller total than 1197. Option C, 26, would give a larger total. Option D, 28, would give an even larger total. A quick check is that the last row is \\(24+17\times5=109\\), and the average of first and last rows is \\(\frac{24+109}{2}=66.5\\); multiplying by 18 gives 1197. Memory cue: in a sum problem, use total, number of terms and difference to find the first term.

The sales form the decreasing AP (220,206,192,\ldots) and (S_{12}=1716). Exam tip: keep (d) negative for decreasing sales.

This is an arithmetic progression with first term 40 and common difference 20. The number of pages read in the nth month is \(a_n=40+(n-1)\times20\). For 300 pages, \(40+(n-1)\times20=300\). Thus, \((n-1)\times20=260\), so \(n-1=13\) and \(n=14\). Therefore, the student will read 300 pages in the 14th month. In the 13th month, the student would read only 280 pages. Exam tip: use the nth-term formula here, not the formula for the sum of terms.

The distances are (25,30,35,\ldots) and (S_{18}=1215). Exam tip: take the number of gaps as the number of terms.

The member numbers are (30,38,46,\ldots) and (S_{15}=1290). Exam tip: treat monthly new members as AP terms.

The passenger numbers are (12,19,26,\ldots) and (S_{16}=1032). Exam tip: take the number of stops as the number of terms.

The profits form the AP (75000,84000,93000,\ldots) and (S_7=714000). Exam tip: treat yearly profits as separate terms.