Word problems based on APs

The fares are (80,100,120,\ldots), and (S_{10}=1700). Exam tip: treat the fare for each kilometre as one term.

The matchstick numbers are (12,19,26,\ldots), and (a_{18}=131). Exam tip: take the figure number as the term number.

The push-ups form the AP (20,25,30,\ldots), and (S_{15}=825). Exam tip: use the sum formula for total practice.

Using (S_{12}=4380) and (a=200) gives (d=30). Exam tip: find the common difference from total and first term.

The governing concept is the nth-term formula for an arithmetic progression. Hourly production forms the sequence 150, 140, 130, and so on, so the first term is a = 150 and the common difference is d = -10 because production decreases by 10 each hour. For the nth hour, a_n = a + (n - 1)d. Setting the production equal to 60 gives 60 = 150 + (n - 1)(-10). Thus 60 = 150 - 10n + 10 = 160 - 10n, so 10n = 100 and n = 10. Therefore option D is correct. Options A, B and C represent earlier hours and give 90, 80 and 70 packets respectively, not 60.

The instrument numbers are (15,20,25,\ldots), and (S_{12}=510). Exam tip: use the sum formula for total installation.

The seats form the AP (30,36,42,\ldots), and (S_{18}=1458). Exam tip: convert row arrangements into APs.

The page numbers are (50,75,100,\ldots), and (a_9=250). Exam tip: use the (n)th term for a particular month.

The sack numbers are (40,37,34,\ldots), and (S_{12}=282). Exam tip: keep the common difference negative for decreasing rows.

The video numbers are (6,9,12,\ldots), and (S_{15}=405). Exam tip: take the number of weeks as the number of terms.

The capacities are (28,32,36,\ldots), and (S_{16}=928). Exam tip: add all rows for total capacity.

The call numbers are (120,135,150,\ldots), and (S_8=1380). Exam tip: treat calls per hour as AP terms.

The weekly sales form an arithmetic progression with first term 60 and common difference 10. Therefore, sales in the nth week are \(a_n=60+(n-1)\times10\). Putting \(a_n=150\), we get \(60+10(n-1)=150\), so \(n-1=9\) and \(n=10\). Hence, the farmer will sell 150 kg in the 10th week. In the 9th week, the sale would be only 140 kg. Exam tip: to find the position of a specified term, equate the target value to \(a_n\).

The admissions are (25,30,35,\ldots), and (S_{20}=1450). Exam tip: use the sum formula for total admissions.

The doses form the AP (5,7,9,\ldots), and (S_{14}=252). Exam tip: add the terms for the total dose.

The numbers of tiles form a decreasing AP with first term \(a=80\) and common difference \(d=-5\). The number of tiles in the \(n\)th row is \(a_n=80+(n-1)(-5)\). From \(80-5(n-1)=25\), we get \(n-1=11\), so \(n=12\). Hence, the 12th row has 25 tiles. The 11th row has 30 tiles, so it is a close but incorrect option. Exam tip: use a negative common difference when quantities decrease.

Using (S_{10}=7750) and (d=150) gives (a=100). Exam tip: the first term can be found from total and difference.

Putting (S_n=504) in the sum formula gives (n=12). Exam tip: you can also test the options in the sum formula.

The distances form the AP (12,15,18,\ldots), and (S_{10}=255). Exam tip: treat each hour's distance as a term.

The plate numbers are (9,13,17,\ldots), and (S_{13}=429). Exam tip: treat each table as one term of the AP.