Word problems based on APs

The distances travelled each hour form an AP: 40, 45, 50, 55, 60. Therefore, \(S_5=40+45+50+55+60=250\) km. 260 km would result from using an incorrect hourly distance or number of terms. In exams, list the first five terms before adding them to avoid mistakes.

The daily sales form an arithmetic progression (AP), with first term 20 and common difference 4. The 11th term is \(a_{11}=a+(11-1)d=20+10\times4=60\). Therefore, 60 boxes are sold on the 11th day. Choosing 56 would add the increase only 9 times, but there are 10 increases from the first day to the 11th day. Exam tip: use \(a_n=a+(n-1)d\) for the nth term.

The sack counts form an arithmetic progression: 10, 15, 20, ... . Here, \(a=10\), \(d=5\), and \(n=7\). Thus, \(S_7=\frac{7}{2}[2(10)+(7-1)5]=\frac{7}{2}(50)=175\). Therefore, there are 175 sacks on 7 racks. Getting 170 usually results from an error while adding the terms or finding the last term. Exam tip: for a total, use the \(S_n\) formula, not just the \(n\)th-term formula.

The gaps form an arithmetic progression with first term \(a=6\) m and common difference \(d=2\) m. The ninth gap is \(a_9=a+(9-1)d=6+8\times2=22\) m. Choosing \(24\) m would incorrectly add 9 increases instead of 8. Exam tip: for the \(n\)th term, use \(n-1\) common differences.

The daily numbers of drawings form an AP: 4, 6, 8, ... . Here, \(a=4\), \(d=2\), and \(n=10\). Therefore, \(S_{10}=\frac{10}{2}[2(4)+(10-1)2]=5(26)=130\). Hence, 130 drawings are made in 10 days. A value such as 120 results from using an incorrect increase or number of days. Exam tip: In AP word problems, first identify \(a\), \(d\), and \(n\), then apply the formula for \(S_n\).

The number of calls forms an arithmetic progression (AP) with first term 35 and common difference 5. The calls handled in the 8th hour are \(a_8=35+(8-1)\times5=35+35=70\). Therefore, 70 is correct. The value 65 results from counting only 6 increases, but there are 7 increases from the first hour to the eighth hour. Exam tip: use \(a_n=a+(n-1)d\) for the nth term of an AP.

The number of pots sold each day forms an arithmetic progression with first term \(a=9\), common difference \(d=3\), and number of terms \(n=12\). Thus, \(S_{12}=\frac{12}{2}[2(9)+(12-1)3]=6(51)=306\). Therefore, 306 pots are sold in 12 days. Option 294 results from not accounting correctly for the daily increase or the number of terms. Exam tip: In AP word problems, first identify \(a\), \(d\), and \(n\), then apply the formula for \(S_n\).

In option B, the amounts are 100, 125, 150, 175, …, so the difference between consecutive terms is always ₹25. An AP requires a constant difference. Option A has a constant ratio, not a constant difference. Exam tip: compare consecutive differences.

The numbers of seats form an AP with first term \(a=14\), common difference \(d=4\), and number of rows \(n=10\). Thus, \(S_{10}=\frac{10}{2}[2(14)+(10-1)\times4]=5(28+36)=320\). Therefore, there are 320 seats in 10 rows. The option 330 can result from using an incorrect number of terms or common difference. Exam tip: In AP word problems, identify \(a\), \(d\), and \(n\) before applying the formula for \(S_n\).

The daily number of words forms an arithmetic progression with first term \(a=10\) and common difference \(d=5\). On the \(8\)th day, \(a_8=a+(8-1)d=10+7\times5=45\). Option 40 counts only 6 increases, whereas reaching the 8th term requires 7 increases. Exam tip: use \(a_n=a+(n-1)d\) for the \(n\)th term of an AP.

From 15 to 18, 18 to 21, and 21 to 24, the increase is 3 each time. A constant difference identifies an AP. A “three times” rule would describe a geometric pattern. In exams, compare consecutive differences first.

Each successive row increases by 3 seats: 11-8=3 and 14-11=3 . Since the common difference is constant, it is an AP. A GP has a constant ratio instead. Exam tip: compare consecutive differences first.

The daily numbers form an arithmetic progression with first term \(a=20\), common difference \(d=4\), and number of terms \(n=9\). Thus, \(S_9=\frac{9}{2}[2(20)+(9-1)\times4]=\frac{9}{2}(72)=324\). Therefore, 324 questions are solved in 9 days. Choosing 330 may result from an error while including the last day’s questions. Exam tip: In AP word problems, identify \(a\), \(d\), and \(n\) before applying the formula for \(S_n\).

The numbers of passengers boarding form an arithmetic progression: 7, 9, 11, \(\ldots\). Here, the first term is \(a=7\), the common difference is \(d=2\), and \(n=10\). Thus, \(a_{10}=a+(n-1)d=7+(10-1)\times2=25\). Therefore, 25 passengers board at the 10th stop. Getting 23 usually results from using the term number incorrectly. Exam tip: for the \(n\)th term of an AP, use \(a_n=a+(n-1)d\).

The plate counts form an arithmetic progression: 6, 9, 12, \ldots. Here, the first term is \(a=6\), the common difference is \(d=3\), and the number of terms is \(n=8\). Thus, \(S_8=\frac{8}{2}[2(6)+(8-1)(3)]=4(12+21)=132\). Therefore, (132) is correct. A value such as (128) can result from using an incorrect common difference or number of terms. Exam tip: In AP word problems, identify \(a\), \(d\), and \(n\) before applying the sum formula.

The costs form an arithmetic progression (AP) with first term \(a=1200\) and common difference \(d=300\). The cost in the sixth year is \(a_6=a+(6-1)d=1200+5\times300=2700\) rupees. Choosing \(3000\) would mean adding the increase 6 times, but there are only 5 increases after the first year. Exam tip: Use \(a_n=a+(n-1)d\) for the \(n\)th term.

The daily numbers form an arithmetic progression: \(5, 8, 11, \ldots\), where \(a=5\), \(d=3\), and \(n=7\). Thus, \(S_7=\frac{7}{2}[2(5)+(7-1)(3)]=\frac{7}{2}(28)=98\). Therefore, 98 computers are installed in 7 days. Choosing 100 does not give the correct sum of the terms with common difference 3. Exam tip: In AP word problems, identify \(a\), \(d\), and \(n\) before applying the formula for \(S_n\).

The daily sales form an arithmetic progression (AP), with first term \(a=80\) and common difference \(d=10\). The sales on the ninth day are \(a_9=a+(9-1)d=80+8\times10=160\). Getting \(170\) would incorrectly add the increase 9 times; there are only 8 increases after the first day. Exam tip: use \(a_n=a+(n-1)d\) for the \(n\)th term.

The daily data amounts form an AP: \(1, 3, 5, \ldots\), where \(a=1\), \(d=2\), and \(n=10\). Thus, \(S_{10}=\frac{10}{2}[2(1)+(10-1)\times2]=5(20)=100\) GB. Therefore, 100 GB is correct. A value such as 95 GB can result from an error while counting terms or applying the sum formula. Exam tip: In AP word problems, identify \(a\), \(d\), and \(n\) before using the formula for \(S_n\).

The daily distances form an AP with first term 20 and common difference 5. The distance on the seventh day is \(a_7=a+(7-1)d=20+6\times5=50\) km. Choosing 55 km would incorrectly count the increase seven times instead of six. Exam tip: use \(a_n=a+(n-1)d\) for the nth term of an AP.

The daily numbers of participants form an arithmetic progression: 30, 40, 50, 60, 70. Therefore, the total for 5 days is 30 + 40 + 50 + 60 + 70 = 250. Option 240 gives a smaller total than the required sum. Exam tip: In AP word problems, first identify the first term, common difference, and number of terms before finding the sum.

This is an arithmetic progression with first term \(a=8\), common difference \(d=2\), and \(n=12\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{12}=8+(12-1)\times2=8+22=30\). Hence, 30 toys are made on the 12th day. Adding \(12\times2\) to 8 would incorrectly give 32, because the increase occurs only 11 times after the first day. Exam tip: use \((n-1)d\), not \(nd\), for the \(n\)th term of an AP.

The numbers of employees joining each month form an AP: \(4, 7, 10, 13, 16, 19\). Here, \(a=4\), \(d=3\), and \(n=6\). Therefore, \(S_6=\frac{6}{2}[2(4)+(6-1)3]=3(23)=69\). Hence, 69 employees join in total. Option 72 can result from using the number of terms or the common difference incorrectly. Exam tip: In AP word problems, identify \(a\), \(d\), and \(n\) before applying the formula for \(S_n\).

The doses form an arithmetic progression with first term \(a=10\) mg and common difference \(d=5\) mg. The dose on day \(n\) is \(a_n=a+(n-1)d\). Thus, \(a_6=10+(6-1)\times5=35\) mg. Option A results from counting too few increases. Exam tip: In the \(n\)th-term formula of an AP, the number of increases is always \(n-1\).

The number of members joining each day forms an AP: 12, 16, 20, ... . Here, a = 12, d = 4, and n = 8. Therefore, S_8 = \(\frac{8}{2}[2(12)+(8-1)4]\) = \(4(24+28)\) = 208. Hence, 208 is correct. Choosing 200 usually results from not accounting correctly for the last term or the number of terms. Exam tip: For total-sum questions, first identify a, d, and n, then apply the \(S_n\) formula.