Word problems based on APs

The distances travelled each hour form an AP: \(55, 60, 65, 70, 75, 80\). Here, \(a=55\), \(d=5\), and \(n=6\). Thus, \(S_6=\frac{6}{2}[2(55)+(6-1)5]=3(135)=405\) km. Therefore, 405 is correct. The value 390 is less than the correct sum of the first six terms. Exam tip: In AP word problems, identify the first term, common difference, and number of terms before applying \(S_n\).

The sales increase by 4 boxes each day, so they form an arithmetic progression. Here, \(a=28\), \(d=4\), and \(n=10\). Thus, \(a_{10}=a+(n-1)d=28+(10-1)\times4=64\). Therefore, 64 boxes are sold on the 10th day. The option \(68\) results from incorrectly adding 10 increases; there are only 9 increases from the first day to the 10th day. Exam tip: use \(n-1\) in the formula for the \(n\)th term of an AP.

The numbers of sacks form an AP with first term \(a=12\), common difference \(d=6\), and \(n=8\) terms. Thus, \(S_8=\frac{8}{2}[2(12)+(8-1)6]=4(66)=264\). Therefore, the total on 8 racks is 264 sacks. Option 288 can result from adding terms incorrectly. Exam tip: for a total, use the sum formula \(S_n\), not only the nth-term formula.

The successive gaps form an arithmetic progression with first term \(a=8\) m and common difference \(d=3\) m. The seventh gap is \(a_7=a+(7-1)d=8+6\times3=26\) m. Choosing \(23\) m would give the fifth gap, not the seventh. Exam tip: in \(a_n=a+(n-1)d\), use \(n-1\), not \(n\).

The daily number of drawings forms an arithmetic progression: \(5, 7, 9, \ldots\), where \(a=5\), \(d=2\), and \(n=12\). Thus, \(S_{12}=\frac{12}{2}[2(5)+(12-1)2]=6(32)=192\). Therefore, 192 is the correct answer. A value such as 190 can result from counting the last term or the number of days incorrectly. Exam tip: in AP word problems, identify \(a\), \(d\), and \(n\) before applying the formula for \(S_n\).

The number of calls handled per hour forms an arithmetic progression with first term 45 and common difference 5. For the ninth hour: \(a_9 = 45 + (9-1)\times 5 = 45 + 40 = 85\). Therefore, 85 is correct. Choosing 90 would result from using the term position incorrectly. Exam tip: use \(a_n=a+(n-1)d\) to find the nth term of an AP.

The number of pots sold each day forms an arithmetic progression with first term \(a=14\), common difference \(d=3\), and \(n=10\) days. Thus, \(S_{10}=\frac{10}{2}[2(14)+(10-1)(3)] = 5(28+27)=275\). Therefore, 275 is the correct answer. A common error leading to 280 is counting the terms or common differences incorrectly. Exam tip: for the total of an AP, use \(S_n=\frac{n}{2}[2a+(n-1)d]\).

The daily amount forms an AP with first term 80 and common difference 20. The 8th term is \(a_8=80+(8-1)\times20=80+140=220\). Therefore, 220 rupees are received on the 8th day. Choosing 210 would not account for the seven increases of 20 from day 1 to day 8. Exam tip: Use \(a_n=a+(n-1)d\) to find the nth term of an AP.

The numbers of rooms form an arithmetic progression: 18, 20, 22, \ldots Here, the first term is \(a=18\), the common difference is \(d=2\), and the number of floors is \(n=9\). Thus, \(S_9=\frac{9}{2}[2(18)+(9-1)(2)]=\frac{9}{2}(52)=234\). Therefore, there are 234 rooms in 9 floors. The value 238 is not obtained by adding the rooms on all nine floors. Exam tip: In AP word problems, identify \(a\), \(d\), and \(n\) before applying the formula for \(S_n\).

This is an arithmetic progression with first term \(a=12\) and common difference \(d=4\). The nth term is \(a_n=a+(n-1)d\). Therefore, \(a_9=12+(9-1)\times4=12+32=44\). Hence, 44 is correct. Choosing 42 would result from incorrectly counting the terms or the daily increase. Exam tip: use \(n-1\), not \(n\), in the nth-term formula.

The numbers of trees planted each day form an AP with first term \(a=18\) and common difference \(d=6\). The seventh term is \(a_7=a+(7-1)d=18+6\times6=54\). Therefore, the correct answer is 54. Getting 48 would result from counting the number of increases incorrectly. Exam tip: use \(n-1\), not \(n\), when finding the \(n\)th term of an AP.

This is an arithmetic progression with first term \(a=16\), common difference \(d=4\), and number of terms \(n=8\). Therefore, \(S_8=\frac{8}{2}[2(16)+(8-1)4]=4(32+28)=240\). Hence, 240 questions are solved in 8 days. A value such as 232 usually results from using the common difference or number of terms incorrectly. Exam tip: identify \(a\), \(d\), and \(n\) before applying the AP sum formula.

This is an arithmetic progression with first term \(a=9\), common difference \(d=3\), and term number \(n=9\). Thus, \(a_9=a+(n-1)d=9+(9-1)\times3=33\). Therefore, 33 passengers board at the 9th stop. Option 36 results from adding 3 nine times, but there are only 8 increases from the first stop to the ninth stop. Exam tip: use \(n-1\) differences when finding the \(n\)th term of an AP.

The plate counts 10, 12, 14, … form an arithmetic progression with a=10, d=2, and n=12. Hence, \(S_{12}=\frac{12}{2}[2(10)+(12-1)(2)] = 6(42)=252\). Therefore, 252 plates are placed on 12 tables. The value 264 can result from using the number of terms or the last term incorrectly. Exam tip: In AP word problems, first identify a, d, and n, then apply the formula for \(S_n\).

The service costs form an arithmetic progression with first term \(a=1500\) and common difference \(d=400\). The fifth-year cost is \(a_5=a+(5-1)d=1500+4\times400=3100\) rupees. Choosing 2900 would count only three increases, but there are four intervals from the first year to the fifth year. Exam tip: Use \(a_n=a+(n-1)d\) and count the intervals carefully.

The numbers of computers installed each day form an AP: \(6, 10, 14, 18, 22, 26\). Here, \(a=6\), \(d=4\), and \(n=6\). Therefore, \(S_6=\frac{6}{2}[2(6)+(6-1)4]=3(32)=96\). Hence, 96 is correct. A value such as 108 can result from using an incorrect number of terms or last term. Exam tip: In word problems, identify the first term, common difference, and number of days before applying the AP sum formula.

The daily sales form an AP with first term \(a=90\) and common difference \(d=15\). The 7th term is \(a_7=a+(7-1)d=90+6\times15=180\). Therefore, 180 is correct. Choosing 195 would correspond to adding the common difference one extra time. Exam tip: for the \(n\)th term, add the common difference \(n-1\) times.

The daily data amounts form an AP with first term \(a=2\), common difference \(d=3\), and \(n=8\) terms. Thus, \(S_8=\frac{8}{2}[2(2)+(8-1)3]=4(25)=100\) GB. Therefore, 100 GB is correct. A value such as 96 GB does not correctly account for the increase of 3 GB on each successive day. Exam tip: In AP word problems, first identify \(a\), \(d\), and \(n\), then use the formula for \(S_n\).

The distances travelled each day form an AP with first term 25 km and common difference 5 km. The eighth term is \(a_8=a+(8-1)d=25+7\times5=60\) km. Therefore, the correct answer is 60 km. Choosing 55 km would count the increase for only 6 days. Exam tip: for the \(n\)th term, multiply the common difference by \(n-1\).

The daily number of participants forms an AP: 40, 50, 60, 70, 80, 90. Here, the first term is \(a=40\), the common difference is \(d=10\), and the number of terms is \(n=6\). Thus, \(S_6=\frac{6}{2}[2(40)+(6-1)(10)]=3(130)=390\). Therefore, 390 is correct. A result such as 400 can arise from counting the terms or days incorrectly. Exam tip: identify \(a\), \(d\), and \(n\) before applying the formula for \(S_n\).

This forms an arithmetic progression with first term \(a=10\) and common difference \(d=3\). The 11th term is \(a_{11}=a+(11-1)d=10+10\times3=40\). Therefore, the correct answer is 40. Getting 42 would result from counting the number of increases incorrectly. Exam tip: Use \(a_n=a+(n-1)d\) for the \(n\)th term of an AP.

The numbers of employees joining each month form an AP: 5, 7, 9, … . Here, a = 5, d = 2, and n = 9. Therefore, S₉ = n/2[2a + (n−1)d] = 9/2[2(5) + 8(2)] = 9/2(26) = 117. Hence, 117 is correct. Exam tip: When a word problem asks for the ‘total’, find the sum of the first n terms, not only the nth term.

The doses form an arithmetic progression (AP), with first term \(a=15\) mg and common difference \(d=5\) mg. The dose on day \(7\) is \(a_7=a+(7-1)d=15+6\times5=45\) mg. Choosing 40 mg counts only 5 increases, but there are 6 increases from day 1 to day 7. Exam tip: Use \(a_n=a+(n-1)d\) for the \(n\)th term of an AP.

The daily number of new members forms an AP: \(15, 20, 25, \ldots\), where \(a=15\), \(d=5\), and \(n=7\). Thus, \(S_7=\frac{7}{2}[2(15)+(7-1)5]=\frac{7}{2}(60)=210\). Therefore, 210 members join in 7 days. The value 195 is the sum for only the first 6 days. Exam tip: When a word problem asks for a total in an AP, use the sum formula \(S_n\), not just the \(n\)th-term formula.

In an AP, the difference between consecutive terms is constant. In option A, the daily difference is ₹20, so \(a_{n+1}-a_n=20\). Option B has a constant ratio and is a GP. Exam tip: check differences first.