Word problems based on APs

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.

The numbers of pots form an arithmetic progression with first term \(a=13\), common difference \(d=2\), and \(n=11\). \(a_{11}=a+(n-1)d=13+(11-1)\times2=33\). Therefore, the 11th section has 33 pots. Choosing 31 would mean adding 2 only nine times. Exam tip: for the \(n\)th term, use \(n-1\) differences, not \(n\).

Write three consecutive AP terms as \(a-d, a, a+d\). Their sum is \(3a=60\), so the middle term is \(a=20\). The first and third terms depend on \(d\). Exam tip: the sum of three consecutive AP terms is always three times the middle term.

The bottle counts form an arithmetic progression with first term \(a=18\) and common difference \(d=3\). For the ninth rack, \(a_9=a+(9-1)d=18+8\times3=42\). Therefore, 42 is correct. Option 45 would result from adding 3 nine times instead of eight times. Exam tip: in the \(n\)th term, add the common difference \(n-1\) times.

The monthly savings form an AP with \(a=50\), \(d=10\), and \(n=8\). Thus, \(S_8=\frac{8}{2}[2(50)+(8-1)(10)]=4(170)=680\). Therefore, the total saving is \(680\) rupees. The value \(640\) results from incorrectly taking the eighth month's saving as \(110\) instead of \(120\). Exam tip: In AP word problems, first identify \(a\), \(d\), and \(n\), then apply the formula for \(S_n\).