Word problems based on APs

Correct answer: B. 80 metres

The daily sales form an arithmetic progression (AP), with first term \(a=35\) and common difference \(d=5\). Sales on the 10th day are \(a_{10}=a+(10-1)d=35+9\times5=80\) metres. Choosing \(85\) metres incorrectly counts 10 increases instead of 9. Exam tip: for the \(n\)th term of an AP, add \(n-1\) common differences.

Correct answer: B. 375

The numbers of seats form an AP with first term \(a=24\), common difference \(d=3\), and \(n=10\) terms. Using \(S_n=\frac{n}{2}[2a+(n-1)d]\), we get \(S_{10}=\frac{10}{2}[2(24)+9(3)]=5(75)=375\). Hence, 375 is correct. A value such as 365 can result from an error in using the number of terms or the common difference in the sum formula. Exam tip: check the last term using \((n-1)d\); here, the tenth row has \(51\) seats.

Correct answer: D. 40 L

The amount filled each minute forms an AP with first term \(a=12\) and common difference \(d=4\). The amount filled in the eighth minute is \(a_8=a+(8-1)d=12+7\times4=40\) L. \(36\) L is the amount for the seventh minute, so it is a close but incorrect option. Exam tip: In \(a_n=a+(n-1)d\), use \(n-1\), not \(n\).

Correct answer: C. ₹1020

The monthly savings form an AP with first term \(a=30\), common difference \(d=10\), and \(n=12\) terms. Thus, \(S_{12}=\frac{12}{2}[2(30)+11(10)]=6(170)=1020\). Therefore, her total saving is ₹1020. ₹990 does not correctly account for the increase through the 12th month. Exam tip: In AP word problems, first identify \(a\), \(d\), and \(n\), then use \(S_n=\frac{n}{2}[2a+(n-1)d]\).

Correct answer: A. 306

The numbers of tiles form a decreasing AP: 50, 46, 42, \ldots Here, \(a=50\), \(d=-4\), and \(n=9\). Therefore, \(S_9=\frac{9}{2}[2(50)+8(-4)]=\frac{9}{2}(68)=306\). Hence, the correct answer is 306. A value such as 316 can result from using the decrease or the number of terms incorrectly. Exam tip: always use a negative common difference for a decreasing AP.

Correct answer: A. The difference between every two consecutive terms is 4.

The sequence is 20, 24, 28, ... . Since 24 − 20 = 4 and 28 − 24 = 4, the common difference is 4, so it is an AP. A constant ratio is a property of a GP, not an AP. Exam tip: check consecutive differences to identify an AP.

Correct answer: A. A student walks 2 km on the first day and 0.5 km more on each succeeding day.

In an AP, the difference between consecutive terms is constant. In option A, the distances are 2, 2.5, 3, 3.5, ...; each increase is 0.5 km. Option B has a constant ratio, so it is a GP. Exam tip: compare successive differences.

Correct answer: C. 42

The number of questions increases by 4 in each successive chapter, so it forms an arithmetic progression. Here, \(a=10\), \(d=4\), and \(n=9\). Thus, \(a_9=a+(n-1)d=10+(9-1)\times4=42\). Therefore, the correct answer is 42. Getting 40 would count only 7 increases, whereas there are 8 gaps from the first to the ninth chapter. Exam tip: for the \(n\)th term, always use \(n-1\) common differences.

Correct answer: B. 560

The seat counts form an arithmetic progression with first term \(a=20\), common difference \(d=2\), and number of terms \(n=16\). Thus, \(S_{16}=\frac{16}{2}[2(20)+(16-1)\times2]=8(70)=560\). Therefore, there are 560 seats in 16 rows. An answer such as 540 can result from incorrectly finding the last term or the number of terms. Exam tip: In AP word problems, identify \(a\), \(d\), and \(n\) before applying the formula for \(S_n\).

Correct answer: C. 80

The number of bricks laid each day forms an AP with first term 40 and common difference 8. The sixth term is \(a_6=a+(6-1)d=40+5\times8=80\). Therefore, the worker will lay 80 bricks on the sixth day. Note that 72 is the number of bricks on the fifth day. Exam tip: Use \(a_n=a+(n-1)d\) to avoid counting the first day incorrectly.

Correct answer: C. 1170

This is an arithmetic progression with first term \(a=120\), common difference \(d=30\), and number of terms \(n=6\). Thus, \(S_6=\frac{6}{2}[2(120)+(6-1)30]=3(390)=1170\). Therefore, a total of 1170 users join in 6 days. The option 1200 can result from an addition or multiplication error. Exam tip: In AP word problems, first identify \(a\), \(d\), and \(n\), then apply the formula for \(S_n\).

Correct answer: B. (35)

The numbers of windows form an arithmetic progression with first term 8 and common difference 3. By the 10th floor, the increase occurs 9 times: \(a_{10}=8+(10-1)\times3=8+27=35\). Therefore, (35) is correct. (38) results from incorrectly adding 3 ten times. Exam tip: in the \(n\)th term of an AP, use \(n-1\) increases.

Correct answer: A. 424

This is an arithmetic progression with first term \(a=32\), common difference \(d=6\), and \(n=8\) terms. Thus, \(S_8=\frac{8}{2}[2(32)+7(6)]=4(64+42)=424\). Therefore, the farmer sells 424 kg of fruit in 8 weeks. Getting 432 usually results from incorrectly counting the number of differences or terms. Exam tip: use \(n-1\) in the AP sum formula.

Correct answer: B. 44

The numbers of admissions form an arithmetic progression with first term \(a=20\) and common difference \(d=4\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_7=20+(7-1)\times4=20+24=44\). Hence, 44 is correct. A common error is to count only 5 increases and obtain 40; there are 6 day-to-day increases from day 1 to day 7. Exam tip: use \(n-1\), not \(n\), when finding the \(n\)th term of an AP.

Correct answer: B. 162

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 \(a_n\).

Correct answer: B. 800 rupees

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.

Correct answer: B. 260

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.

Correct answer: C. 44

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.

Correct answer: C. 392

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.

Correct answer: B. 26

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.

Correct answer: C. 260

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 \(S_n\).

Correct answer: B. 65 litres

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.

Correct answer: B. 245

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 \(S_n\).

Correct answer: B. 300

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.

Correct answer: B. 140

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.