Word problems based on APs

The weekly numbers of uploaded videos form the AP 6, 9, 12, \(\ldots\), where \(a=6\), \(d=3\), and \(n=9\). Thus, \(S_9=\frac{9}{2}[2(6)+(9-1)3]=\frac{9}{2}(36)=162\). Therefore, 162 videos are uploaded in 9 weeks. The value 168 can result from using an incorrect term or common difference. Exam tip: for a total in AP word problems, use the sum formula \(S_n\), not just the nth-term formula \(a_n\).

The monthly deposits form an arithmetic progression (AP) with first term \(a=300\) and common difference \(d=100\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_6=300+(6-1)\times100=800\) rupees, so option B is correct. Choosing 700 rupees counts only four increases, but there are five increases from the first month to the sixth month. Exam tip: always use \((n-1)\) in the formula for the \(n\)th term.

The row capacities form an AP: 15, 20, 25, ... . Here, \(a=15\), \(d=5\), and \(n=8\). Therefore, \(S_8=\frac{8}{2}[2(15)+(8-1)5]=4(65)=260\). Hence, the total capacity is 260 cars. A result such as 270 comes from an error in counting the terms or applying the common difference. Exam tip: identify \(a\), \(d\), and \(n\) before using the AP sum formula.

The numbers of dots form an arithmetic progression with first term \(a=4\) and common difference \(d=5\). The ninth term is \(a_9=a+(9-1)d=4+8\times5=44\). Therefore, 44 is correct. Option 42 results from counting the increases incorrectly. Exam tip: for the \(n\)th term, remember that there are \(n-1\) intervals after the first term.

This forms an arithmetic progression with first term \(a=35\), common difference \(d=7\), and number of terms \(n=7\). Therefore, \(S_7=\frac{7}{2}[2(35)+(7-1)\times7]=\frac{7}{2}(112)=392\). Hence, 392 books are arranged in 7 days. The value 406 can result from an error in counting the terms or common difference. Exam tip: identify \(a\), \(d\), and \(n\) before applying the AP sum formula.

The marks form an arithmetic progression with first term \(a=4\), common difference \(d=2\), and \(n=12\). Thus, \(a_{12}=a+(n-1)d=4+(12-1)\times2=26\). The option 28 results from incorrectly using 12 increases instead of the 11 gaps before the 12th term. Exam tip: in the \(n\)th-term formula, add the common difference \(n-1\) times.

This is an arithmetic progression with first term \(a=22\), common difference \(d=3\), and number of terms \(n=8\). The total number of patients is \(S_8=\frac{8}{2}[2(22)+(8-1)(3)] = 4(65)=260\). A value such as 264 can result from using the common difference or number of days incorrectly. Exam tip: In AP word problems, first identify \(a\), \(d\), and \(n\), then apply the formula for \(S_n\).

The amount filled each hour forms an arithmetic progression with first term \(a=25\) and common difference \(d=5\). The amount in the 9th hour is \(a_9=a+(9-1)d=25+8\times5=65\) litres. Choosing 60 litres counts only 7 increases. Exam tip: for the \(n\)th term, add the common difference \(n-1\) times.

The numbers of flowers form an arithmetic progression with first term \(a=11\), common difference \(d=3\), and \(n=10\) rows. Thus, \(S_{10}=\frac{10}{2}[2(11)+(10-1)\times3]=5(22+27)=245\). Therefore, 245 is correct. A value such as 240 can result from using an incorrect last term or number of terms. Exam tip: In AP word problems, first identify \(a\), \(d\), and \(n\), then apply the formula for \(S_n\).

The daily production increases uniformly by 30 toys, so it forms an arithmetic progression. Here, \(a=180\), \(d=30\), and \(n=5\). Therefore, \(a_5=a+(5-1)d=180+4\times30=300\). Option 270 adds the increase only three times, but four increases occur by the fifth day. Exam tip: use \(a+(n-1)d\) for the \(n\)th term.

The numbers of students form an arithmetic progression: \(8, 12, 16, \ldots\), where \(a=8\), \(d=4\), and \(n=7\). Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), we get \(S_7=\frac{7}{2}[2(8)+6(4)]=\frac{7}{2}(40)=140\). Therefore, the correct answer is 140. The option 144 may result from adding the terms incorrectly. Exam tip: identify \(a\), \(d\), and \(n\) before applying the AP sum formula.

The distances travelled each hour form an AP: \(55, 60, 65, 70, 75, 80\). Here, \(a=55\), \(d=5\), and \(n=6\). Thus, \(S_6=\frac{6}{2}[2(55)+(6-1)5]=3(135)=405\) km. Therefore, 405 is correct. The value 390 is less than the correct sum of the first six terms. Exam tip: In AP word problems, identify the first term, common difference, and number of terms before applying \(S_n\).

The sales increase by 4 boxes each day, so they form an arithmetic progression. Here, \(a=28\), \(d=4\), and \(n=10\). Thus, \(a_{10}=a+(n-1)d=28+(10-1)\times4=64\). Therefore, 64 boxes are sold on the 10th day. The option \(68\) results from incorrectly adding 10 increases; there are only 9 increases from the first day to the 10th day. Exam tip: use \(n-1\) in the formula for the \(n\)th term of an AP.

The numbers of sacks form an AP with first term \(a=12\), common difference \(d=6\), and \(n=8\) terms. Thus, \(S_8=\frac{8}{2}[2(12)+(8-1)6]=4(66)=264\). Therefore, the total on 8 racks is 264 sacks. Option 288 can result from adding terms incorrectly. Exam tip: for a total, use the sum formula \(S_n\), not only the nth-term formula.

The successive gaps form an arithmetic progression with first term \(a=8\) m and common difference \(d=3\) m. The seventh gap is \(a_7=a+(7-1)d=8+6\times3=26\) m. Choosing \(23\) m would give the fifth gap, not the seventh. Exam tip: in \(a_n=a+(n-1)d\), use \(n-1\), not \(n\).

The daily number of drawings forms an arithmetic progression: \(5, 7, 9, \ldots\), where \(a=5\), \(d=2\), and \(n=12\). Thus, \(S_{12}=\frac{12}{2}[2(5)+(12-1)2]=6(32)=192\). Therefore, 192 is the correct answer. A value such as 190 can result from counting the last term or the number of days incorrectly. Exam tip: in AP word problems, identify \(a\), \(d\), and \(n\) before applying the formula for \(S_n\).

The number of calls handled per hour forms an arithmetic progression with first term 45 and common difference 5. For the ninth hour: \(a_9 = 45 + (9-1)\times 5 = 45 + 40 = 85\). Therefore, 85 is correct. Choosing 90 would result from using the term position incorrectly. Exam tip: use \(a_n=a+(n-1)d\) to find the nth term of an AP.

The number of pots sold each day forms an arithmetic progression with first term \(a=14\), common difference \(d=3\), and \(n=10\) days. Thus, \(S_{10}=\frac{10}{2}[2(14)+(10-1)(3)] = 5(28+27)=275\). Therefore, 275 is the correct answer. A common error leading to 280 is counting the terms or common differences incorrectly. Exam tip: for the total of an AP, use \(S_n=\frac{n}{2}[2a+(n-1)d]\).

The daily amount forms an AP with first term 80 and common difference 20. The 8th term is \(a_8=80+(8-1)\times20=80+140=220\). Therefore, 220 rupees are received on the 8th day. Choosing 210 would not account for the seven increases of 20 from day 1 to day 8. Exam tip: Use \(a_n=a+(n-1)d\) to find the nth term of an AP.

The numbers of rooms form an arithmetic progression: 18, 20, 22, \ldots Here, the first term is \(a=18\), the common difference is \(d=2\), and the number of floors is \(n=9\). Thus, \(S_9=\frac{9}{2}[2(18)+(9-1)(2)]=\frac{9}{2}(52)=234\). Therefore, there are 234 rooms in 9 floors. The value 238 is not obtained by adding the rooms on all nine floors. Exam tip: In AP word problems, identify \(a\), \(d\), and \(n\) before applying the formula for \(S_n\).