Word problems based on APs

Correct answer: C. (22)

Using (S_9=414) and (d=6) gives (a=22). Exam tip: the first term can be found from total and common difference.

Correct answer: C. (23)

Putting (S_n=1610) in the sum formula gives (n=23). Exam tip: when total seats are given, form an equation for (n).

Correct answer: B. (12)

The amounts form an AP and (S_n=2550) gives (n=12). Exam tip: set the target total equal to (S_n).

Correct answer: D. (2280)

The production is the decreasing AP (300,280,260,\ldots) and (S_{12}=2280). Exam tip: treat a decrease as a negative common difference.

Correct answer: C. (5)

Using (S_{15}=795) and (a=18) gives (d=5). Exam tip: find the common difference from the total and first term.

Correct answer: C. 17

The numbers of plants form an arithmetic progression with first term 16 and common difference 6. Thus, the number of plants in the nth row is \(a_n=16+(n-1)\times6\). On putting \(16+(n-1)\times6=112\), we get \(n-1=16\), so \(n=17\). Therefore, the 17th row has 112 plants. The 16th row would have only 106 plants. Exam tip: in AP word problems, identify the first term and common difference, then equate the target value to \(a_n\).

Correct answer: C. (2320)

The daily work forms the AP (40,48,56,\ldots) and (S_{20}=2320). Exam tip: treat the daily increase as the common difference.

Correct answer: C. (10)

The donation amounts are (150,225,300,\ldots) and (S_n=4875) gives (n=10). Exam tip: form an (S_n) equation for the target total.

Correct answer: C. 12th row

The numbers of tiles, counted from the bottom, form a decreasing AP with first term 110 and common difference -5. Thus, the number of tiles in the nth row is \(110+(n-1)(-5)\). Setting this equal to 55 gives \(110-5(n-1)=55\), so \(n-1=11\) and \(n=12\). Therefore, the 12th row has 55 tiles. The 11th row has 60 tiles, so it is a close but incorrect option. Exam tip: use a negative common difference when the terms decrease upwards.

Correct answer: B. (1908)

The fares are (60,78,96,\ldots) and (S_{12}=1908). Exam tip: treat the fare for each kilometre as one term.

Correct answer: C. (130)

The matchstick numbers are (10,18,26,\ldots) and (a_{16}=130). Exam tip: take the figure number as the term number.

Correct answer: C. (7140)

The deposits are (250,290,330,\ldots) and (S_{14}=7140). Exam tip: treat weekly deposits as AP terms.

Correct answer: C. (12750)

The amounts form the decreasing AP (1200,1150,1100,\ldots) and (S_{15}=12750). Exam tip: keep (d) negative for decreasing quantities.

Correct answer: B. (43500)

The salary forms an AP and (a_{10}=43500). Exam tip: find a particular month's salary using the (n)th term.

Correct answer: B. (1701)

The capacities are (35,42,49,\ldots) and (S_{18}=1701). Exam tip: add all rows for total capacity.

Correct answer: C. (981)

The pot numbers are (12,17,22,\ldots) and (S_{18}=981). Exam tip: use the sum formula for total production.

Correct answer: C. (20)

The admissions are (30,40,50,\ldots) and (S_n=2500) gives (n=20). Exam tip: connect the target total with the sum formula.

Correct answer: B. (236)

The hourly production forms an arithmetic progression because the machine adds the same number of units, 12, every hour. Here the first term is 80, the common difference is 12, and the 14th hour means the 14th term. The term formula is \\(a_n=a+(n-1)d\\), so \\(a_{14}=80+(14-1)12=80+156=236\\).

Therefore, option B, 236, is correct. It is important not to multiply 14 directly by 12, because the first hour already starts at 80 and only the next 13 hours add an increase. The sequence begins 80, 92, 104 and continues with a constant difference of 12. Treating the hour number as the term number gives the required production.

Correct answer: C. (520)

The work hours are (16,20,24,\ldots) and (S_{13}=520). Exam tip: add all days using the sum formula for total hours.

Correct answer: C. (53000)

The instalments are (4000,4750,5500,\ldots) and (S_8=53000). Exam tip: treat instalments as AP terms.

Correct answer: B. (1320)

The sales form the decreasing AP (180,168,156,\ldots) and (S_{11}=1320). Exam tip: keep (d) negative for decreasing sales.

Correct answer: C. 15

The numbers of pages form an arithmetic progression with first term \(a=25\) and common difference \(d=15\). The number of pages read in the \(n\)th month is \(a_n=a+(n-1)d\). So, \(25+(n-1)\times15=235\). This gives \((n-1)\times15=210\), hence \(n-1=14\) and \(n=15\). Therefore, the student reads 235 pages in the 15th month. In the 14th month, the student would read only 220 pages. Exam tip: for a specified term of an AP, use \(a_n=a+(n-1)d\).

Correct answer: C. (966)

The distances are (30,36,42,\ldots) and (S_{14}=966). Exam tip: take the number of gaps as the number of terms.

Correct answer: C. (726)

The member numbers are (22,29,36,\ldots) and (S_{12}=726). Exam tip: treat monthly new members as AP terms.

Correct answer: C. (1007)

The passenger numbers are (8,13,18,\ldots) and (S_{19}=1007). Exam tip: take the number of stops as the number of terms.