Word problems based on APs

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.

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.