Word problems based on APs

The number of plants on the first day is 6, and 4 more plants are added each subsequent day. Thus, the terms are 6, 6 + 4 = 10, 10 + 4 = 14, and 14 + 4 = 18. Therefore, the sequence is \(6,10,14,18,\ldots\), with common difference 4. Option B is incorrect because its first term is 4 instead of 6. Exam tip: To identify an AP, check whether the difference between consecutive terms remains constant.

The direct answer is C: 360 seats. The rows form an arithmetic progression because each row has a fixed increase of 2 seats. Here, first term is an actual number of seats, so use a = 10, common difference d = 2, and number of rows n = 15. The last row is a + (n-1)d = 10 + 14×2 = 38. The total is S_n = n/2[2a + (n-1)d] = 15/2[20 + 28] = 15/2×48 = 360. Option A, 340, is too small; it can result from an incorrect addition or from using the wrong last term. Option B, 350, is also an arithmetic error. Option C matches the formula and calculation. Option D, 370, is too large. Exam cue: for total rows use the sum formula, not only the last-term formula.

The governing concept is the sum of the first n terms of an arithmetic progression. Ravi’s daily savings are 20, 25, 30, and so on, so the first term is a = 20, the common difference is d = 5, and n = 10. The tenth saving is a + 9d = 20 + 45 = 65 rupees. Hence the total is Sₙ = n/2[2a + (n − 1)d] = 10/2[40 + 45] = 5 × 85 = 425 rupees. Therefore option B is correct. The other choices can result from using an incorrect number of increases, confusing the tenth-day amount with the total, or adding the terms inaccurately. The fixed difference confirms that an AP-sum method is appropriate.

The numbers of plants form an AP with first term \(a=6\), common difference \(d=3\), and \(n=8\) terms. \(S_8=\frac{8}{2}[2(6)+(8-1)3]=4(12+21)=132\). Therefore, 132 plants are planted in 8 rows. A value such as 120 may result from an incorrect addition up to the eighth row. Exam tip: when the total of all terms is asked, use the sum formula \(S_n\), not only the nth-term formula \(a_n\).

The numbers of pages form an arithmetic progression with first term \(a=5\) and common difference \(d=2\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{12}=5+(12-1)\times2=5+22=27\). Hence, the student reads 27 pages on the 12th day. Choosing 25 would count only 10 increases, but there are 11 increases from the first term to the 12th term. Exam tip: use \((n-1)\), not \(n\), when finding the \(n\)th term of an AP.

The numbers of bricks form a decreasing AP: \(30, 28, 26, \ldots\). Here, \(a=30\), \(d=-2\), and \(n=10\). The 10th row has \(a_{10}=30+9(-2)=12\) bricks. Therefore, \(S_{10}=\frac{10}{2}(30+12)=210\). Hence, 210 is correct. The value 220 would result from incorrectly taking 14 as the last term; 14 is the 9th term. Exam tip: find the first and last terms, then use \(S_n=\frac{n}{2}(a+l)\).

The numbers of passengers boarding at successive stops form an AP: \(4, 7, 10, \ldots\). Here, \(a=4\), \(d=3\), and \(n=9\). Thus, \(S_9=\frac{9}{2}[2(4)+(9-1)(3)]=\frac{9}{2}(32)=144\). Therefore, the correct answer is 144. A value such as 136 results from an incorrect count of terms or common difference. Exam tip: when a word problem asks for the total, use \(S_n\), not just the \(n\)th-term formula.

The running times 15, 20, 25, ... form an arithmetic progression (AP), with first term \(a=15\) and common difference \(d=5\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_7=15+(7-1)\times5=45\) minutes. Choosing 50 minutes would give the time for the eighth day, not the seventh. Exam tip: in the \(n\)th-term formula of an AP, add the common difference \(n-1\) times.

The daily notebook sales form the AP \(12,16,20,24,28,32\). Here, \(a=12\), \(d=4\), and \(n=6\). Thus, \(S_6=\frac{6}{2}[2(12)+(6-1)4]=3(44)=132\). Therefore, the correct answer is 132. Option 128 may result from omitting the last term or making an addition error. Exam tip: In AP word problems, first identify \(a\), \(d\), and \(n\), then apply the formula for \(S_n\).

The amount filled each minute forms an arithmetic progression. Here, the first term is \(a=8\) and the common difference is \(d=2\). Thus, \(a_{10}=a+(10-1)d=8+9\times2=26\). Therefore, \(26\ \text{L}\) of water is filled in the 10th minute. \(24\ \text{L}\) would be the amount filled in the 8th minute, not the 10th. Exam tip: use \(a_n=a+(n-1)d\) to find the \(n\)th term of an AP.

The matchstick counts form an arithmetic progression with first term \(a=3\) and common difference \(d=4\). The 11th term is \(a_{11}=a+(11-1)d=3+10\times4=43\). Option 41 can result from incorrectly counting the number of increases. Exam tip: always use \(a_n=a+(n-1)d\) for the nth term.

The numbers of books arranged each day form an AP: 25, 30, 35, \ldots Here, the first term is \(a=25\), the common difference is \(d=5\), and the number of days is \(n=8\). Thus, \(S_8=\frac{8}{2}[2(25)+(8-1)5]=4(50+35)=340\). Therefore, 340 books are arranged in 8 days. A value such as 350 may result from incorrectly adding the last-day count; for a total, use the \(S_n\) formula. Exam tip: In AP word problems, identify \(a\), \(d\), and \(n\) first.

The weekly sales form an arithmetic progression with first term \(a=40\) kg and common difference \(d=10\) kg. Therefore, the fifth-week sale is \(a_5=a+(5-1)d=40+4\times10=80\) kg. Choosing \(90\) kg incorrectly adds the increase five times; there are only four increases after the first week. Exam tip: use \(a_n=a+(n-1)d\) for the \(n\)th term.

The seat numbers form an arithmetic progression with first term \(a=18\) and common difference \(d=2\). The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(a_{12}=18+(12-1)\times2=18+22=40\). Therefore, the 12th row has 40 seats. The answer 42 results from incorrectly adding \(12\times2\) instead of \(11\times2\). Exam tip: always use \((n-1)\) in the formula for the \(n\)th term.

The numbers of words learned each day form an AP with first term \(a=7\), common difference \(d=3\), and \(n=9\) terms. Therefore, \(S_9=\frac{9}{2}[2(7)+(9-1)\times3]=\frac{9}{2}(38)=171\). Hence, the child learns 171 words in 9 days. A value such as 180 results from an incorrect count of terms or common difference. Exam tip: when the total is asked, use the sum formula \(S_n\), not just the nth-term formula.

The row capacities form an AP: \(20,24,28,\ldots\), where \(a=20\), \(d=4\), and \(n=7\). Thus, \(S_7=\frac{7}{2}[2(20)+(7-1)4]=\frac{7}{2}(64)=224\). Therefore, the total capacity is 224 cars. A value such as 232 can result from using an incorrect last term. Exam tip: In AP word problems, identify \(a\), \(d\), and \(n\) before applying the sum formula.

The number of bricks laid each day forms an arithmetic progression with first term \(a=50\) and common difference \(d=10\). The sixth term is \(a_6=a+(6-1)d=50+5\times10=100\). Therefore, the correct answer is 100. Note that 90 is the number of bricks laid on the fifth day. Exam tip: Use \(a_n=a+(n-1)d\) for the nth term of an AP.

The daily donations form an arithmetic progression with first term \(a=100\), common difference \(d=20\), and \(n=7\) terms. Therefore, \(S_7=\frac{7}{2}[2(100)+(7-1)(20)]=\frac{7}{2}(320)=1120\) rupees. Hence, 1120 is correct. An option such as 1100 may result from using the increase or the number of days incorrectly. Exam tip: identify \(a\), \(d\), and \(n\) before applying the formula for \(S_n\).

The numbers of flowers in the layers form an AP: \(5,10,15,\ldots\). Here, \(a=5\), \(d=5\), and \(n=10\). Thus, \(S_{10}=\frac{10}{2}[2(5)+(10-1)5]=5(55)=275\). Therefore, there are 275 flowers in 10 layers. The value 250 can result from an incorrect sum; in AP questions, always check the number of terms \(n\).

This is an arithmetic progression with first term \(a=2\) GB and common difference \(d=1\) GB. The nth term is \(a_n=a+(n-1)d\). Therefore, \(a_{14}=2+(14-1)\times1=15\) GB. Option A incorrectly ignores that the increase is added 13 times after the first day. Exam tip: For the nth term of an AP, use \(n-1\) common differences.

The numbers of watches repaired each hour form the AP \(6,8,10,\ldots\), where \(a=6\), \(d=2\), and \(n=8\). Therefore, \(S_8=\frac{8}{2}[2(6)+(8-1)(2)]=4(12+14)=104\). Hence, the correct answer is \(104\). Finding only the work done in the eighth hour would not give the total work. Exam tip: In AP word problems, use \(S_n\) for a total, not \(a_n\).

The water amounts form a decreasing arithmetic progression, with first term \(a=1000\) and common difference \(d=-50\). On the eighth day, \(a_8=a+(8-1)d=1000+7(-50)=650\) litres. Choosing \(600\) litres would subtract 50 eight times, but the given amount is already for day 1. Exam tip: use \(a+(n-1)d\) for the \(n\)th term.

This is an AP because the number of seats increases by the same fixed difference, 3, in every row; for example, 14−11=3. In a GP, consecutive terms have a constant ratio instead. Exam tip: subtract adjacent terms to identify an AP.

The daily production forms an AP with first term 200 and common difference 25. The production on the fifth day is \(a_5=a+(5-1)d=200+4\times25=300\). Adding 25 five times gives 325, but only four increases occur after the first day. Exam tip: use \(a_n=a+(n-1)d\) for the \(n\)th term of an AP.

The numbers of students form an AP: 30, 35, 40, 45, 50, 55. Here, \(a=30\), \(d=5\), and \(n=6\). Therefore, \(S_6=\frac{6}{2}[2(30)+(6-1)5]=3(85)=255\). Hence, 255 is the correct answer. 260 can result from incorrectly adding the final class size. Exam tip: identify \(a\), \(d\), and \(n\) before applying the AP sum formula.