Word problems based on APs

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.

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.

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 \(n\).

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 \(S_n=\frac{n}{2}[2a+(n-1)d]\).

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.

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.

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.

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.

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 \(S_n\).

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.

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 \(S_n\).

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.

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.

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.

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.