Word problems based on APs

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.

The savings form the AP (80,100,120,\ldots) and (S_{12}=2280). Exam tip: use the sum formula for total amount.

The seats form an AP and (a_{20}=18+19(3)=75). Exam tip: take the row number as (n) and find the (n)th term.

The ticket numbers are (40,45,50,\ldots) and (S_{15}=1125). Exam tip: identify (a), (d), and (n) before finding total sales.

This is the decreasing AP (52,48,44,\ldots) and (S_{10}=340). Exam tip: treat the decrease as a negative common difference.

The plant numbers are (10,14,18,\ldots) and (S_{12}=384). Exam tip: take the total number of rows as (n).

The salary forms an AP and (a_8=12000+7(500)=15500). Exam tip: find the monthly salary using the (n)th term.

The practice times form the AP (25,30,35,\ldots) and (S_{14}=805). Exam tip: the same sum formula works when time increases regularly.

The donation amounts are (200,250,300,\ldots) and (S_{10}=4250). Exam tip: first and last day amounts are useful for total donation.

The matchstick numbers are (6,10,14,\ldots) and (a_{15}=62). Exam tip: in pattern questions check whether the difference is constant.

The book numbers are (22,28,34,\ldots) and (S_9=414). Exam tip: take the number of shelves as the number of terms.

The production is the AP (150,175,200,\ldots) and (S_7=1575). Exam tip: treat the daily increase as the common difference.

The distances are (10,12,14,\ldots) and (S_{11}=220). Exam tip: the number of gaps and the number of poles are different.

The marks form the AP (2,5,8,\ldots) and (S_{12}=222). Exam tip: take the number of questions as (n) and find the sum.

The direct answer is C: there are 16 rows. The numbers of chairs form an arithmetic progression because each row increases by the same amount, 2. Here the first term is a=12, common difference is d=2, and the total is S_n=432. Use the sum formula: S_n=\frac{n}{2}[2a+(n-1)d]. Substitution gives 432=\frac{n}{2}[24+2(n-1)]=\frac{n}{2}(2n+22)=n(n+11). Thus n^2+11n-432=0. Factoring gives (n-16)(n+27)=0, so n=16 because a number of rows cannot be negative. Option A, 14, gives 14(25)=350 chairs, not 432. Option B, 15, gives 15(26)=390. Option C, 16, gives 16(27)=432 and is correct. Option D, 17, gives 17(28)=476. The common mistake is to use the last term formula without checking the total. First identify the AP, then use the sum formula.

The direct answer is B, 225 litres. The amounts form a decreasing arithmetic progression because the same quantity, 25 litres, is removed every day. The first term is \\(a=500\\), the common difference is \\(d=-25\\), and the required term is the 12th term. There are 11 decreases before day 12, so \\(a_{12}=a+(12-1)d=500+11(-25)=500-275=225\\). Option B is therefore correct. Option A, 200, would result from using too many decreases or making another arithmetic error. Option C, 250, is not the result after eleven reductions; it is too large. Option D, 275, also ignores part of the required decrease. The minus sign is essential because the quantity goes down, not up. Memory cue: for the nth term, use first term plus \\(n-1\\) times the common difference; count the changes before the requested day.

Using (S_8=440) and (d=10) gives (a=20). Exam tip: solve the sum formula for the unknown first term.

The direct answer is B, 24 bulbs. This is an arithmetic progression because the number changes by the same amount in every step. The first row has 45 bulbs, and each row above has 3 fewer, so the common difference is -3. The eighth term is found by counting seven changes after the first row: \(a_8=a_1+(8-1)d=45+7(-3)=45-21=24\). Therefore option B is correct. Option A, 21, would result from subtracting 24, which is too much. Option B, 24, follows the rule exactly. Option C, 27, subtracts only 18 and therefore corresponds to the seventh row, not the eighth. Option D, 30, subtracts only 15 and corresponds to the sixth row. Remember: the first row is already counted, so an eighth term has seven differences.

The participant numbers are (16,20,24,\ldots) and (S_9=288). Exam tip: treat each round as one term.

The stickers form the AP (5,10,15,\ldots) and (S_{20}=1050). Exam tip: treat page patterns with a constant difference as APs.

Here, the first term is \(a=1000\), the number of terms is \(n=6\), and the sum is \(S_6=9000\). Using the AP sum formula \(S_n=\frac{n}{2}[2a+(n-1)d]\), we get \(9000=\frac{6}{2}[2(1000)+5d]\). Thus, \(9000=6000+15d\), so \(d=200\). Therefore, the monthly increase is 200 rupees. If the increase were 250 rupees, the total for six months would not be 9000 rupees. Exam tip: in questions stating an equal monthly increase, treat that increase as the common difference \(d\).

The seats are (15,19,23,\ldots) and (S_{18}=882). Exam tip: you may also find the last row and use (S_n=\frac{n}{2}(a+l)).

Putting (a_n=700), (100+(n-1)50=700), gives (n=13). Exam tip: use the (n)th term for a particular month's amount.