Word problems based on APs

The situation forms an arithmetic progression because the load changes by the same amount, 15 kg, after every cycle. The first load is a = 240 and the common difference is d = −15. Let the load in cycle n be 75 kg. Applying aₙ = a + (n − 1)d gives 75 = 240 − 15(n − 1). Thus 15(n − 1) = 240 − 75 = 165. Dividing by 15 gives n − 1 = 11, so n = 12. Therefore option C, the 12th cycle, is correct. A direct check confirms the result: cycle 10 has 240 − 9(15) = 105 kg, cycle 11 has 90 kg, cycle 12 has 75 kg, and cycle 13 has 60 kg. The negative difference must be retained because the load decreases rather than increases.

This is an arithmetic-progression word problem. The first-stage value is a₁ = 1250, and a decrease of 55 at every next stage means the common difference is d = −55. Thus the value at stage n is aₙ = a₁ + (n − 1)d = 1250 − 55(n − 1). Set this equal to 260: 1250 − 55(n − 1) = 260. Rearranging gives 55(n − 1) = 990, so n − 1 = 18 and n = 19. Checking directly, after 18 decreases the value is 1250 − 18 × 55 = 1250 − 990 = 260. Therefore the value is 260 at the 19th stage, so option C is correct. The negative common difference is essential because the sequence decreases.

The number solved each day forms an arithmetic progression: 64, 79, 94, …, with first term a₁ = 64 and common difference d = 15. We are looking for the daily number, so we use the nth-term formula rather than the sum formula. Set aₙ = 514: 514 = 64 + (n − 1)15. Subtracting 64 gives 450 = 15(n − 1). Dividing by 15 gives n − 1 = 30, hence n = 31. A direct check gives the 31st-day value as 64 + 30 × 15 = 514. Therefore, option C, the 31st day, is correct. The nearby choices arise if one counts the first day incorrectly or confuses the number solved on a day with the cumulative total solved over several days.

Putting (S_n=1350) in the sum formula gives (n=25). Exam tip: when total seats are given, form an equation for (n).

The amounts form the AP (40,55,70,\ldots), and (S_n=1470) gives (n=12). Exam tip: set the target total equal to (S_n).

The salaries form the AP (25000,26200,27400,\ldots), and (S_8=233600). Exam tip: add all monthly salaries using the sum formula.

This is the decreasing AP (96,90,84,\ldots), and (S_{12}=756). Exam tip: treat a decrease as a negative common difference.

Using (S_{10}=500) and (a=14) gives (d=8). Exam tip: find the common difference from the total and first term.

The plant numbers are (12,16,20,\ldots), and (S_{16}=672). Exam tip: take the number of rows as the number of terms.

The passenger numbers are (5,9,13,\ldots), and (a_{15}=61). Exam tip: treat the stop number as the (n)th term.

The daily work forms the AP (32,44,56,\ldots), and (S_{14}=1540). Exam tip: treat the daily increase as the common difference.

The donation amounts are (100,150,200,\ldots), and (S_n=3250) gives (n=10). Exam tip: form an (S_n) equation for the target total.

This forms an arithmetic progression with first term 24 and common difference 5. Therefore, the number of seats in the nth row is \(a_n=24+(n-1)\times5\). Setting \(24+(n-1)\times5=99\) gives \(5(n-1)=75\), so \(n=16\). Hence, the 16th row has 99 seats. The 15th row would have only 94 seats. Exam tip: for row-number questions, first write \(a_n=a+(n-1)d\).

The tile numbers are (7,13,19,\ldots), and (S_{14}=644). Exam tip: find the total of a pattern using the sum formula.

The distances are (20,25,30,\ldots), and (S_{16}=920). Exam tip: take the number of gaps as the number of terms.

The amounts form the decreasing AP (1000,960,920,\ldots), and (S_{10}=8200). Exam tip: keep (d) negative for decreasing quantities.

The profits form the AP (50000,55000,60000,\ldots), and (S_8=540000). Exam tip: treat yearly profits as separate terms.

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 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.