Word problems based on APs

The sales form the decreasing AP (56,52,48,\ldots) and (S_9=360). Exam tip: in decreasing sales, (d) is negative.

Using (S_{10}=650) and (a=20) gives (d=10). Exam tip: find the common difference from the total and first term.

The number of computers is (10,14,18,\ldots) and (S_{12}=384). Exam tip: use the sum formula for total installation.

The daily data amounts form an arithmetic progression because the amount increases by the same quantity, 3 GB, each day. Here the first term is a = 5 and the common difference is d = 3. To find the amount on the 16th day, use the nth-term formula a_n = a + (n - 1)d. Substituting the values gives a_16 = 5 + (16 - 1) × 3 = 5 + 15 × 3 = 5 + 45 = 50 GB. Hence option B is correct. Option A results from using only 14 increases, while option C adds one extra 3-GB increase; option D reflects two extra increases. The key is that the first day is the first term, so only 15 increases occur before day 16.

The cloth sales form (70,82,94,\ldots) and (S_8=896). Exam tip: use the sum formula for sales across all days.

This forms an arithmetic progression with first term 12 and common difference 5. Hence, \(a_n=12+(n-1)\times5\). Putting \(a_n=77\), we get \(12+5(n-1)=77\), so \(5(n-1)=65\), \(n-1=13\), and \(n=14\). Therefore, 77 instruments will be placed on the 14th day. On the 13th day, there would be only 72 instruments. Exam tip: in AP word problems, identify the first term and common difference, then equate the target value to \(a_n\).

Putting (S_n=1350) in the sum formula gives (n=25). Exam tip: when total seats are given, form an equation for (n).

The amounts form the AP (40,55,70,\ldots), and (S_n=1470) gives (n=12). Exam tip: set the target total equal to (S_n).

The salaries form the AP (25000,26200,27400,\ldots), and (S_8=233600). Exam tip: add all monthly salaries using the sum formula.

This is the decreasing AP (96,90,84,\ldots), and (S_{12}=756). Exam tip: treat a decrease as a negative common difference.

Using (S_{10}=500) and (a=14) gives (d=8). Exam tip: find the common difference from the total and first term.

The plant numbers are (12,16,20,\ldots), and (S_{16}=672). Exam tip: take the number of rows as the number of terms.

The passenger numbers are (5,9,13,\ldots), and (a_{15}=61). Exam tip: treat the stop number as the (n)th term.

The daily work forms the AP (32,44,56,\ldots), and (S_{14}=1540). Exam tip: treat the daily increase as the common difference.

The donation amounts are (100,150,200,\ldots), and (S_n=3250) gives (n=10). Exam tip: form an (S_n) equation for the target total.

This forms an arithmetic progression with first term 24 and common difference 5. Therefore, the number of seats in the nth row is \(a_n=24+(n-1)\times5\). Setting \(24+(n-1)\times5=99\) gives \(5(n-1)=75\), so \(n=16\). Hence, the 16th row has 99 seats. The 15th row would have only 94 seats. Exam tip: for row-number questions, first write \(a_n=a+(n-1)d\).

The tile numbers are (7,13,19,\ldots), and (S_{14}=644). Exam tip: find the total of a pattern using the sum formula.

The distances are (20,25,30,\ldots), and (S_{16}=920). Exam tip: take the number of gaps as the number of terms.

The amounts form the decreasing AP (1000,960,920,\ldots), and (S_{10}=8200). Exam tip: keep (d) negative for decreasing quantities.

The profits form the AP (50000,55000,60000,\ldots), and (S_8=540000). Exam tip: treat yearly profits as separate terms.