Word problems based on APs

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.

The daily numbers of instruments form an arithmetic progression: 7, 10, 13, \ldots. Here, the first term is \(a=7\), the common difference is \(d=3\), and the number of terms is \(n=10\). Therefore, \(S_{10}=\frac{10}{2}[2(7)+(10-1)(3)]=5(14+27)=205\). Hence, 205 instruments are placed in 10 days. An answer such as 200 usually results from using an incorrect number of intervals or an incorrect value of the common difference. Exam tip: identify \(a\), \(d\), and \(n\) before applying the AP sum formula.

The savings form the AP (80,100,120,\ldots) and (S_{12}=2280). Exam tip: use the sum formula for total amount.

The seats form an AP and (a_{20}=18+19(3)=75). Exam tip: take the row number as (n) and find the (n)th term.

The ticket numbers are (40,45,50,\ldots) and (S_{15}=1125). Exam tip: identify (a), (d), and (n) before finding total sales.