Word problems based on APs

Correct answer: B. 250 km

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.

Correct answer: B. 60

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.

Correct answer: C. 175

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.

Correct answer: B. 22 m

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.

Correct answer: B. 130

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 \(S_n\).

Correct answer: B. 70

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.

Correct answer: B. 306

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 \(S_n\).

Correct answer: B. ₹100 is received on the first day, and ₹25 more is received each following day.

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.

Correct answer: B. 320

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 \(S_n\).

Correct answer: B. 45

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.

Correct answer: A. The difference between consecutive terms is constant

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.

Correct answer: A. Arithmetic progression (AP)

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.

Correct answer: C. 324

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 \(S_n\).

Correct answer: B. 25

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 \(a_n=a+(n-1)d\).

Correct answer: B. (132)

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.

Correct answer: B. ₹2,700

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.

Correct answer: A. 98

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 \(S_n\).

Correct answer: C. 160

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.

Correct answer: C. 100 GB

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 \(S_n\).

Correct answer: B. 50 km

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.

Correct answer: B. 250

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.

Correct answer: B. 30

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.

Correct answer: B. 69

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 \(S_n\).

Correct answer: B. 35 mg

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 \(n-1\).

Correct answer: C. 208

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.