Word problems based on APs

The numbers of stars form an arithmetic progression with first term 9 and common difference 3. The 10th term is \(a_{10}=a+(10-1)d=9+9\times3=36\). Therefore, the correct answer is 36. The value 39 would be the 11th term, not the 10th term. Exam tip: use \(a_n=a+(n-1)d\) to find the \(n\)th term of an AP.

The daily practice times form an AP: 30, 40, 50, 60, 70. Hence, the total is \(30+40+50+60+70=250\) minutes. The value 225 would result from incorrectly taking the average as 45 minutes; for five terms, the correct average is the middle term, 50 minutes. Exam tip: For an odd number of AP terms, sum = number of terms × middle term.

The gaps form an AP: \(5,10,15,\ldots\), where \(a=5\), \(d=5\), and \(n=8\). Therefore, \(S_8=\frac{8}{2}[2(5)+(8-1)5]=4(45)=180\) m. Hence, 180 m is correct. Choosing 170 m results from not including the eighth gap correctly. Exam tip: In AP word problems, identify the first term, common difference, and number of terms before applying \(S_n\).

The numbers of boxes form an arithmetic progression with first term \(a=16\), common difference \(d=4\), and number of terms \(n=9\).
\(S_9=\frac{9}{2}[2(16)+(9-1)(4)]\)
\(=\frac{9}{2}[32+32]=\frac{9}{2}\times64=288\).
Therefore, the total number of boxes on 9 shelves is 288. Computing only \(16+8\times4=48\) gives the number on the ninth shelf, not the total. Exam tip: when the question asks for a total, use \(S_n\), not \(a_n\).

The lap lengths form an arithmetic progression with first term 100 and common difference 20. The sixth term is \(a_6=a+(6-1)d=100+5\times20=200\) metres. Choosing 220 metres incorrectly adds the increase six times; there are only five increases after the first lap. Exam tip: use \(a_n=a+(n-1)d\) for the nth term of an AP.

The monthly numbers of new employees form an arithmetic progression: \(3, 5, 7, \ldots\). Here, \(a=3\), \(d=2\), and \(n=12\). Therefore, \(S_{12}=\frac{12}{2}[2(3)+(12-1)\times2]=6(28)=168\). Hence, 168 is the correct answer. A value such as 170 can result from an error in counting the terms or using the common difference. Exam tip: In AP word problems, first identify \(a\), \(d\), and \(n\), then apply the formula for \(S_n\).

The numbers of trees planted each day form an AP with first term \(a=11\) and common difference \(d=4\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{10}=11+(10-1)\times4=11+36=47\). Hence, 47 is correct. Getting 45 would result from using the number of intervals incorrectly. Exam tip: for the \(n\)th term of an AP, multiply the common difference by \((n-1)\).

In option A, consecutive differences are equal: 15−10=5, 20−15=5 and 25−20=5. Hence it is an AP. In option B, the differences are 5, 10 and 20. Exam tip: always check consecutive differences.

The distances covered each hour form an AP: \(12, 14, 16, 18, 20, 22\). Here, \(a=12\), \(d=2\), and \(n=6\). Thus, \(S_6=\frac{6}{2}[2(12)+(6-1)\times2]=3(34)=102\) km. Therefore, 102 km is correct. The value 104 km can result from incorrectly finding the last term or the number of terms. Exam tip: In an AP word problem, identify \(a\), \(d\), and \(n\) before applying the formula for \(S_n\).

The monthly deposits form an arithmetic progression with first term \(a=500\) and common difference \(d=100\). The ninth term is \(a_9=a+(9-1)d=500+8\times100=1300\). Therefore, the correct deposit is ₹1300. ₹1200 would count only 7 increases, whereas there are 8 increases from the first month to the ninth month. Exam tip: use \(a_n=a+(n-1)d\) to find the \(n\)th term of an AP.

The numbers of windows form an AP: 4, 8, 12, ..., where the first term is a = 4, common difference d = 4, and n = 8. The eighth floor has 4 + 7 × 4 = 32 windows. Therefore, S₈ = 8/2 × (4 + 32) = 144 windows. The value 136 would result from incorrectly taking the last term as 30. Exam tip: For AP sum questions, first find the nth term and then use Sₙ = n/2(a + l).

The numbers of students form an arithmetic progression: 5, 10, 15, 20, 25, 30. Here, the first term is \(a=5\), the common difference is \(d=5\), and \(n=6\). Thus, \(a_6=a+(6-1)d=5+5\times5=30\). Therefore, 30 is correct. The value 25 belongs to the fifth group, not the sixth. Exam tip: use \(n-1\) in the formula for the \(n\)th term of an AP.

The plate counts form an arithmetic progression with first term \(a=8\), common difference \(d=2\), and \(n=10\) terms. Thus, \(S_{10}=\frac{10}{2}[2(8)+(10-1)\times2]=5(34)=170\). Therefore, 170 plates are placed on 10 tables. Getting 180 usually results from using an incorrect last term or number of terms. Exam tip: In AP word problems, identify \(a\), \(d\), and \(n\) before applying the formula for \(S_n\).

The distances travelled each day form an AP with first term \(a=15\) km and common difference \(d=5\) km. The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_8=15+(8-1)\times5=15+35=50\) km. Hence, 50 km is correct. Choosing 55 km would give the 9th term, not the 8th term. Exam tip: Use \(n-1\), not \(n\), in the nth-term formula.

The marks form the AP 40, 45, 50, 55, 60, 65. Here, \(a=40\), \(d=5\), and \(n=6\). Thus, \(S_6=\frac{6}{2}[2(40)+(6-1)5]=3(105)=315\). Therefore, 315 is correct. Choosing 320 results from adding the increase incorrectly. Exam tip: An AP with \(n\) terms has \(n-1\) common differences.

The service costs form an AP with first term ₹2000 and common difference ₹500. The cost in the fifth year is \(a_5=a+(5-1)d=2000+4\times500=4000\). Hence, ₹4000 is correct. Choosing ₹4500 incorrectly counts 5 increases; from the first year to the fifth year, there are only 4 increases. Exam tip: use \(a_n=a+(n-1)d\) for the nth term of an AP.

The daily number of drawings forms an arithmetic progression with first term \(a=3\), common difference \(d=1\), and \(n=15\) days. On the 15th day, the artist makes \(a_{15}=3+(15-1)\times1=17\) drawings. Therefore, \(S_{15}=\frac{15}{2}(3+17)=150\) drawings in total. Choosing 155 would add one extra drawing to the sum. Exam tip: in AP word problems, identify \(a\), \(d\), and \(n\) first, then use \(S_n=\frac{n}{2}[2a+(n-1)d]\).

In option A, the number of seats increases by 4 in every new row, so the common difference is constant and it forms an AP. In option B, the number doubles. Exam tip: check the difference between consecutive terms.

The hourly ticket sales form an AP with first term \(a=100\), common difference \(d=50\), and \(n=6\) terms. \(S_6=\frac{6}{2}[2(100)+(6-1)50]=3(450)=1350\). Therefore, 1350 tickets are sold in 6 hours. The value 1300 would fail to include the sixth hour’s sale of 350 tickets. Exam tip: When a word problem asks for a total, use \(S_n\), not just the \(n\)th-term formula.

The number of toys made each day forms an AP with first term \(a=9\) and common difference \(d=3\). The eighth term is \(a_8=a+(8-1)d=9+7\times3=30\). Therefore, the correct answer is 30. Choosing 27 would add the increase only 6 times, whereas there are 7 increases from day 1 to day 8. Exam tip: use \(a_n=a+(n-1)d\), taking care to use \(n-1\), not \(n\).

The numbers of candidates at the desks form an AP: 2, 4, 6, \ldots, 20. Here, \(a=2\), \(d=2\), and \(n=10\). Therefore, \(S_{10}=\frac{10}{2}[2(2)+(10-1)\times2]=5(22)=110\). Hence, 110 candidates can sit at the 10 desks. The option 100 may result from an incorrect addition. Exam tip: in AP word problems, first identify \(a\), \(d\), and \(n\), then apply the formula for \(S_n\).

The sack counts form an arithmetic progression with first term \(a=14\) and common difference \(d=6\). For the \(7\)th rack, \(a_7=a+(7-1)d=14+6\times6=50\). Therefore, 50 is correct. The value 56 would result from adding the difference seven times instead of six times. Exam tip: for the \(n\)th term, add the common difference \(n-1\) times.

The daily patient counts form an arithmetic progression: 10, 12, 14, \ldots. Here, \(a=10\), \(d=2\), and \(n=12\). Therefore, \(S_{12}=\frac{12}{2}[2(10)+11(2)]=6(42)=252\). Hence, the correct answer is 252. The value 264 is the number of patients on the 12th day, \(a_{12}\), not the total over 12 days. Exam tip: distinguish between \(a_n\) for one term and \(S_n\) for the sum of terms.

The numbers of videos watched each day form an AP with first term \(a=4\), common difference \(d=2\), and \(n=9\). Thus, \(a_9=a+(n-1)d=4+(9-1)\times2=20\). Therefore, 20 is correct. The value 18 would represent the 8th-day term, not the 9th-day term. Exam tip: use \(n-1\), not \(n\), in the formula \(a_n=a+(n-1)d\).

The daily sales form an arithmetic progression with first term \(a=25\), common difference \(d=5\), and \(n=7\) terms. Thus, \(S_7=\frac{7}{2}[2(25)+(7-1)5]=\frac{7}{2}(80)=280\). Therefore, \(280\) boxes are sold in 7 days. \(285\) results from using an incorrect number of terms or common difference. Exam tip: identify \(a\), \(d\), and \(n\) before applying the AP sum formula.