Word problems based on APs

The number of plants on the first day is 6, and 4 more plants are added each subsequent day. Thus, the terms are 6, 6 + 4 = 10, 10 + 4 = 14, and 14 + 4 = 18. Therefore, the sequence is \(6,10,14,18,\ldots\), with common difference 4. Option B is incorrect because its first term is 4 instead of 6. Exam tip: To identify an AP, check whether the difference between consecutive terms remains constant.

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.

The governing idea is modelling repeated equal decrease as an arithmetic progression. The first-stage value is a=760, and because the temperature decreases by 32 at every step, the common difference is d=−32. The value at stage n is therefore a_n=760+(n−1)(−32). Set the target equal to 280: 760−32(n−1)=280. Hence 32(n−1)=480, so n−1=15 and n=16. Therefore the temperature is 280 units at the 16th stage, making option C correct. The negative difference must be used because the sequence decreases. Checking nearby stages gives stage 15 as 312 and stage 17 as 248, so neither neighboring option satisfies the target.

The governing concept is an arithmetic progression for a quantity that increases by a fixed amount each day. The first-day number is a=42 and the daily increase is d=9. Since the question asks how many questions are solved on one particular day, use the nth-term formula, not the sum formula: a_n=a+(n−1)d. Set 276=42+(n−1)9. Then 234=9(n−1), so n−1=26 and n=27. Therefore 276 questions are solved on the 27th day, so option C is correct. The sum formula would calculate the total number solved over several days and would answer a different question. The neighboring choices arise from an off-by-one error in counting the first day or the number of increases.

The loads form an arithmetic progression because the same quantity, 45 units, is subtracted at every successive stage. The first term is a₁ = 980 and the common difference is d = −45. Using aₙ = a₁ + (n − 1)d, set the load equal to 305: 305 = 980 − 45(n − 1). Rearranging gives 675 = 45(n − 1), so n − 1 = 15 and n = 16. Therefore the load is 305 units at the 16th stage, making option C correct. The nearby choices arise from counting the number of reductions incorrectly, treating the first stage as a reduction, or using a positive common difference. The negative sign is essential because the load decreases.

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.

This is an AP with (a=10), (d=2), and (n=15). Exam tip: use the sum formula for total seats.

The daily savings form an arithmetic progression because the amount increases by the same fixed difference, 5 rupees, each day. Here the first term is a = 20, the common difference is d = 5, and the number of terms is n = 10. The tenth-day saving is a + 9d = 20 + 9(5) = 65 rupees. Therefore, the total is S_n = n/2[2a + (n − 1)d] = 10/2[40 + 9(5)] = 5(85) = 425 rupees. Hence option B is correct. The other options result from using an incorrect final term or adding the progression inaccurately.

The numbers of plants form an AP with first term \(a=6\), common difference \(d=3\), and \(n=8\) terms. \(S_8=\frac{8}{2}[2(6)+(8-1)3]=4(12+21)=132\). Therefore, 132 plants are planted in 8 rows. A value such as 120 may result from an incorrect addition up to the eighth row. Exam tip: when the total of all terms is asked, use the sum formula \(S_n\), not only the nth-term formula \(a_n\).

The numbers of pages form an arithmetic progression with first term \(a=5\) and common difference \(d=2\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{12}=5+(12-1)\times2=5+22=27\). Hence, the student reads 27 pages on the 12th day. Choosing 25 would count only 10 increases, but there are 11 increases from the first term to the 12th term. Exam tip: use \((n-1)\), not \(n\), when finding the \(n\)th term of an AP.

The numbers of bricks form a decreasing AP: \(30, 28, 26, \ldots\). Here, \(a=30\), \(d=-2\), and \(n=10\). The 10th row has \(a_{10}=30+9(-2)=12\) bricks. Therefore, \(S_{10}=\frac{10}{2}(30+12)=210\). Hence, 210 is correct. The value 220 would result from incorrectly taking 14 as the last term; 14 is the 9th term. Exam tip: find the first and last terms, then use \(S_n=\frac{n}{2}(a+l)\).

The numbers of passengers boarding at successive stops form an AP: \(4, 7, 10, \ldots\). Here, \(a=4\), \(d=3\), and \(n=9\). Thus, \(S_9=\frac{9}{2}[2(4)+(9-1)(3)]=\frac{9}{2}(32)=144\). Therefore, the correct answer is 144. A value such as 136 results from an incorrect count of terms or common difference. Exam tip: when a word problem asks for the total, use \(S_n\), not just the \(n\)th-term formula.

The running times 15, 20, 25, ... form an arithmetic progression (AP), with first term \(a=15\) and common difference \(d=5\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_7=15+(7-1)\times5=45\) minutes. Choosing 50 minutes would give the time for the eighth day, not the seventh. Exam tip: in the \(n\)th-term formula of an AP, add the common difference \(n-1\) times.

The daily notebook sales form the AP \(12,16,20,24,28,32\). Here, \(a=12\), \(d=4\), and \(n=6\). Thus, \(S_6=\frac{6}{2}[2(12)+(6-1)4]=3(44)=132\). Therefore, the correct answer is 132. Option 128 may result from omitting the last term or making an addition error. Exam tip: In AP word problems, first identify \(a\), \(d\), and \(n\), then apply the formula for \(S_n\).

The amount filled each minute forms an arithmetic progression. Here, the first term is \(a=8\) and the common difference is \(d=2\). Thus, \(a_{10}=a+(10-1)d=8+9\times2=26\). Therefore, \(26\ \text{L}\) of water is filled in the 10th minute. \(24\ \text{L}\) would be the amount filled in the 8th minute, not the 10th. Exam tip: use \(a_n=a+(n-1)d\) to find the \(n\)th term of an AP.

The matchstick counts form an arithmetic progression with first term \(a=3\) and common difference \(d=4\). The 11th term is \(a_{11}=a+(11-1)d=3+10\times4=43\). Option 41 can result from incorrectly counting the number of increases. Exam tip: always use \(a_n=a+(n-1)d\) for the nth term.

The numbers of books arranged each day form an AP: 25, 30, 35, \ldots Here, the first term is \(a=25\), the common difference is \(d=5\), and the number of days is \(n=8\). Thus, \(S_8=\frac{8}{2}[2(25)+(8-1)5]=4(50+35)=340\). Therefore, 340 books are arranged in 8 days. A value such as 350 may result from incorrectly adding the last-day count; for a total, use the \(S_n\) formula. Exam tip: In AP word problems, identify \(a\), \(d\), and \(n\) first.

The weekly sales form an arithmetic progression with first term \(a=40\) kg and common difference \(d=10\) kg. Therefore, the fifth-week sale is \(a_5=a+(5-1)d=40+4\times10=80\) kg. Choosing \(90\) kg incorrectly adds the increase five times; there are only four increases after the first week. Exam tip: use \(a_n=a+(n-1)d\) for the \(n\)th term.

The seat numbers form an arithmetic progression with first term \(a=18\) and common difference \(d=2\). The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(a_{12}=18+(12-1)\times2=18+22=40\). Therefore, the 12th row has 40 seats. The answer 42 results from incorrectly adding \(12\times2\) instead of \(11\times2\). Exam tip: always use \((n-1)\) in the formula for the \(n\)th term.