Word problems based on APs

The numbers of words learned each day form an AP with first term \(a=7\), common difference \(d=3\), and \(n=9\) terms. Therefore, \(S_9=\frac{9}{2}[2(7)+(9-1)\times3]=\frac{9}{2}(38)=171\). Hence, the child learns 171 words in 9 days. A value such as 180 results from an incorrect count of terms or common difference. Exam tip: when the total is asked, use the sum formula \(S_n\), not just the nth-term formula.

The row capacities form an AP: \(20,24,28,\ldots\), where \(a=20\), \(d=4\), and \(n=7\). Thus, \(S_7=\frac{7}{2}[2(20)+(7-1)4]=\frac{7}{2}(64)=224\). Therefore, the total capacity is 224 cars. A value such as 232 can result from using an incorrect last term. Exam tip: In AP word problems, identify \(a\), \(d\), and \(n\) before applying the sum formula.

The number of bricks laid each day forms an arithmetic progression with first term \(a=50\) and common difference \(d=10\). The sixth term is \(a_6=a+(6-1)d=50+5\times10=100\). Therefore, the correct answer is 100. Note that 90 is the number of bricks laid on the fifth day. Exam tip: Use \(a_n=a+(n-1)d\) for the nth term of an AP.

The daily donations form an arithmetic progression with first term \(a=100\), common difference \(d=20\), and \(n=7\) terms. Therefore, \(S_7=\frac{7}{2}[2(100)+(7-1)(20)]=\frac{7}{2}(320)=1120\) rupees. Hence, 1120 is correct. An option such as 1100 may result from using the increase or the number of days incorrectly. Exam tip: identify \(a\), \(d\), and \(n\) before applying the formula for \(S_n\).

The numbers of flowers in the layers form an AP: \(5,10,15,\ldots\). Here, \(a=5\), \(d=5\), and \(n=10\). Thus, \(S_{10}=\frac{10}{2}[2(5)+(10-1)5]=5(55)=275\). Therefore, there are 275 flowers in 10 layers. The value 250 can result from an incorrect sum; in AP questions, always check the number of terms \(n\).

This is an arithmetic progression with first term \(a=2\) GB and common difference \(d=1\) GB. The nth term is \(a_n=a+(n-1)d\). Therefore, \(a_{14}=2+(14-1)\times1=15\) GB. Option A incorrectly ignores that the increase is added 13 times after the first day. Exam tip: For the nth term of an AP, use \(n-1\) common differences.

The numbers of watches repaired each hour form the AP \(6,8,10,\ldots\), where \(a=6\), \(d=2\), and \(n=8\). Therefore, \(S_8=\frac{8}{2}[2(6)+(8-1)(2)]=4(12+14)=104\). Hence, the correct answer is \(104\). Finding only the work done in the eighth hour would not give the total work. Exam tip: In AP word problems, use \(S_n\) for a total, not \(a_n\).

The water amounts form a decreasing arithmetic progression, with first term \(a=1000\) and common difference \(d=-50\). On the eighth day, \(a_8=a+(8-1)d=1000+7(-50)=650\) litres. Choosing \(600\) litres would subtract 50 eight times, but the given amount is already for day 1. Exam tip: use \(a+(n-1)d\) for the \(n\)th term.

This is an AP because the number of seats increases by the same fixed difference, 3, in every row; for example, 14−11=3. In a GP, consecutive terms have a constant ratio instead. Exam tip: subtract adjacent terms to identify an AP.

The daily production forms an AP with first term 200 and common difference 25. The production on the fifth day is \(a_5=a+(5-1)d=200+4\times25=300\). Adding 25 five times gives 325, but only four increases occur after the first day. Exam tip: use \(a_n=a+(n-1)d\) for the \(n\)th term of an AP.

The numbers of students form an AP: 30, 35, 40, 45, 50, 55. Here, \(a=30\), \(d=5\), and \(n=6\). Therefore, \(S_6=\frac{6}{2}[2(30)+(6-1)5]=3(85)=255\). Hence, 255 is the correct answer. 260 can result from incorrectly adding the final class size. Exam tip: identify \(a\), \(d\), and \(n\) before applying the AP sum formula.

The numbers of stars form an arithmetic progression with first term 9 and common difference 3. The 10th term is \(a_{10}=a+(10-1)d=9+9\times3=36\). Therefore, the correct answer is 36. The value 39 would be the 11th term, not the 10th term. Exam tip: use \(a_n=a+(n-1)d\) to find the \(n\)th term of an AP.

The daily practice times form an AP: 30, 40, 50, 60, 70. Hence, the total is \(30+40+50+60+70=250\) minutes. The value 225 would result from incorrectly taking the average as 45 minutes; for five terms, the correct average is the middle term, 50 minutes. Exam tip: For an odd number of AP terms, sum = number of terms × middle term.

The gaps form an AP: \(5,10,15,\ldots\), where \(a=5\), \(d=5\), and \(n=8\). Therefore, \(S_8=\frac{8}{2}[2(5)+(8-1)5]=4(45)=180\) m. Hence, 180 m is correct. Choosing 170 m results from not including the eighth gap correctly. Exam tip: In AP word problems, identify the first term, common difference, and number of terms before applying \(S_n\).

The numbers of boxes form an arithmetic progression with first term \(a=16\), common difference \(d=4\), and number of terms \(n=9\).
\(S_9=\frac{9}{2}[2(16)+(9-1)(4)]\)
\(=\frac{9}{2}[32+32]=\frac{9}{2}\times64=288\).
Therefore, the total number of boxes on 9 shelves is 288. Computing only \(16+8\times4=48\) gives the number on the ninth shelf, not the total. Exam tip: when the question asks for a total, use \(S_n\), not \(a_n\).

The lap lengths form an arithmetic progression with first term 100 and common difference 20. The sixth term is \(a_6=a+(6-1)d=100+5\times20=200\) metres. Choosing 220 metres incorrectly adds the increase six times; there are only five increases after the first lap. Exam tip: use \(a_n=a+(n-1)d\) for the nth term of an AP.

The monthly numbers of new employees form an arithmetic progression: \(3, 5, 7, \ldots\). Here, \(a=3\), \(d=2\), and \(n=12\). Therefore, \(S_{12}=\frac{12}{2}[2(3)+(12-1)\times2]=6(28)=168\). Hence, 168 is the correct answer. A value such as 170 can result from an error in counting the terms or using the common difference. Exam tip: In AP word problems, first identify \(a\), \(d\), and \(n\), then apply the formula for \(S_n\).

The numbers of trees planted each day form an AP with first term \(a=11\) and common difference \(d=4\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{10}=11+(10-1)\times4=11+36=47\). Hence, 47 is correct. Getting 45 would result from using the number of intervals incorrectly. Exam tip: for the \(n\)th term of an AP, multiply the common difference by \((n-1)\).

In option A, consecutive differences are equal: 15−10=5, 20−15=5 and 25−20=5. Hence it is an AP. In option B, the differences are 5, 10 and 20. Exam tip: always check consecutive differences.

The distances covered each hour form an AP: \(12, 14, 16, 18, 20, 22\). Here, \(a=12\), \(d=2\), and \(n=6\). Thus, \(S_6=\frac{6}{2}[2(12)+(6-1)\times2]=3(34)=102\) km. Therefore, 102 km is correct. The value 104 km can result from incorrectly finding the last term or the number of terms. Exam tip: In an AP word problem, identify \(a\), \(d\), and \(n\) before applying the formula for \(S_n\).