Word problems based on APs

The numbers of tiles form a decreasing AP: \(40, 37, 34, \ldots\). Here, \(a=40\), \(d=-3\), and \(n=8\). Thus, \(S_8=\frac{8}{2}[2(40)+7(-3)]=4(59)=236\). Therefore, there are 236 tiles in 8 rows. Option 244 can result from using the decrease or the number of terms incorrectly. Exam tip: when terms decrease, use a negative common difference.

The numbers of plants form an arithmetic progression (AP), with first term a = 12 and common difference d = 4. The 7th term is a₇ = a + (7 − 1)d = 12 + 6 × 4 = 36. Therefore, 36 is correct. Getting 34 results from an incorrect count of the increases or common difference. Exam tip: always use aₙ = a + (n − 1)d for the nth term.

The daily distances form an AP with first term \(a=2\), common difference \(d=1\), and number of terms \(n=10\). Therefore, \(S_{10}=\frac{10}{2}[2(2)+(10-1)(1)]=5(13)=65\) km. Hence, 65 km is correct. A total of 60 km may result from incorrectly taking the distance on the 10th day as 10 km. Exam tip: In AP word problems, identify \(a\), \(d\), and \(n\) before applying the formula for \(S_n\).

The tile counts 5, 7, 9, ... form an AP. Its last term is 23, so \(S_{10}=\frac{10}{2}(5+23)=140\). Ravi ignores the increase in later rows. Exam tip: find the last term before adding an AP.

The seat counts form an arithmetic progression with first term 16, common difference 2, and 20 terms. Thus, \(S_{20}=\frac{20}{2}[2(16)+(20-1)\times2]=10(32+38)=700\). Therefore, there are 700 seats in 20 rows. An answer such as 720 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 sum formula.

The number of bricks laid each day forms an AP with first term \(a=30\) and common difference \(d=6\). For the ninth day, \(a_9=a+(9-1)d=30+8\times6=78\). Therefore, the correct answer is (78). (84) would result from adding 6 nine times to 30, but there are only 8 increases after the first day. Exam tip: use \((n-1)d\) in the formula for the \(n\)th term.

The numbers of users joining each day form an AP: \(100, 150, 200, 250, 300\). Here, \(a=100\), \(d=50\), and \(n=5\). Thus, \(S_5=\frac{5}{2}[2(100)+(5-1)50]=1000\). Therefore, 1000 users join in total in 5 days. Choosing 1050 does not give the correct sum of the five terms. Exam tip: Check whether the question asks for the total \(S_n\) or only the \(n\)th term.

The numbers of windows form an AP with first term \(a=6\) and common difference \(d=2\). For the 12th floor, \(a_{12}=a+(12-1)d=6+11\times2=28\). Therefore, 28 is correct. Choosing 26 would incorrectly use 10 increases; there are 11 increases from the first floor to the 12th floor. Exam tip: use \(n-1\) in the formula for the \(n\)th term.

The weekly sales form an arithmetic progression with first term \(a=25\), common difference \(d=5\), and \(n=10\) terms. Thus, \(S_{10}=\frac{10}{2}[2(25)+(10-1)5]=5(95)=475\) kg. Therefore, option B is correct. Taking 500 kg does not account for the weekly increase correctly. Exam tip: In AP word problems, first identify \(a\), \(d\), and \(n\).

The admissions form an arithmetic progression with first term \(a=15\) and common difference \(d=3\). Admissions on the sixth day are \(a_6=a+(6-1)d=15+5\times3=30\). \(33\) would be the number of admissions on the seventh day, so it is a close but incorrect option. Exam tip: Use \(n-1\), not \(n\), when finding the \(n\)th term of an AP.

The weekly numbers of videos form the AP \(4, 6, 8, \ldots\). Here, \(a=4\), \(d=2\), and \(n=8\). Therefore, \(S_8=\frac{8}{2}[2(4)+(8-1)\times2]=4(8+14)=88\). Hence, 88 videos are uploaded in 8 weeks. Option 86 can result from an addition error; use the sum-of-first-\(n\)-terms formula for an AP. Exam tip: identify \(a\), \(d\), and \(n\) before substituting values.

The monthly deposits form an arithmetic progression (AP), with first term \(a=200\) and common difference \(d=50\). The seventh-month deposit is \(a_7=a+(7-1)d=200+6\times 50=500\). Choosing \(₹550\) would incorrectly add seven increases instead of six. Exam tip: in the \(n\)th term of an AP, the number of increases is always \(n-1\).

The row capacities form the AP \(12, 16, 20, 24, 28, 32\). Here, \(a=12\), \(d=4\), and \(n=6\). Therefore, \(S_6=\frac{6}{2}[2(12)+(6-1)4]=3(44)=132\). Hence, the total capacity is 132 cars. Option 136 can 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 numbers of dots form an arithmetic progression: 5, 8, 11, \(\ldots\). Here, the first term is \(a=5\), the common difference is \(d=3\), and \(n=10\). Therefore, \(a_{10}=a+(n-1)d=5+(10-1)\times3=32\). Hence, 32 is correct. Adding \(10\times3\) to 5 would incorrectly give 35, because the increase occurs only 9 times after the first figure. Exam tip: use \(n-1\), not \(n\), in the AP nth-term formula.

The daily numbers of books form an AP: 40, 48, 56, 64, 72, 80. Here, \(a=40\), \(d=8\), and \(n=6\). Therefore, \(S_6=\frac{6}{2}[2(40)+(6-1)8]=3(120)=360\). Hence, 360 is correct. A value such as 380 results from using the sum formula or the number of terms incorrectly. Exam tip: identify \(a\), \(d\), and \(n\) before applying the AP sum formula.

The marks form an arithmetic progression with first term \(a=3\) and common difference \(d=1\). The \(n\)th term is \(a_n=a+(n-1)d\). Hence, \(a_{15}=3+(15-1)\times1=17\). Therefore, the 15th question carries 17 marks. Choosing 18 would incorrectly add \(15d\); the required gap is \(15-1=14\). Exam tip: in the formula for the \(n\)th term, use \(n-1\), not \(n\).

The daily number of patients forms an arithmetic progression: \(18, 20, 22, \ldots\). Here, \(a=18\), \(d=2\), and \(n=9\). Therefore, \(S_9=\frac{9}{2}[2(18)+(9-1)2]=\frac{9}{2}(52)=234\). Hence, 234 is correct. An answer such as 230 may result from using the wrong number of terms or making an error in the sum formula. Exam tip: identify \(a\), \(d\), and \(n\) before applying the AP formula.

The amounts filled each hour form an AP with first term \(a=30\) and common difference \(d=10\). The amount filled in the eighth hour is \(a_8=a+(8-1)d=30+7\times10=100\) litres. Getting 110 litres would incorrectly add nine differences instead of seven. Exam tip: for the \(n\)th term, use \(n-1\) common differences.

The flower counts form an arithmetic progression with first term \(a=7\), common difference \(d=4\), and \(n=9\) terms. Thus, \(S_9=\frac{9}{2}[2(7)+(9-1)4]=\frac{9}{2}(46)=207\). Therefore, there are 207 flowers in 9 rows. A value such as 216 can result from incorrectly treating a row value as the total. Exam tip: when a word problem asks for the total, use \(S_n\), not only \(a_n\).

The daily production forms an AP with first term \(a=150\) and common difference \(d=20\). Thus, the production on the sixth day is \(a_6=a+(6-1)d=150+5\times20=250\). The answer 270 results from adding 20 six times, but there are only five increases after the first day. Exam tip: use \(a_n=a+(n-1)d\) for the \(n\)th term.