Word problems based on APs

The tree numbers are (25,30,35,\ldots), and (S_n=900) gives (n=15). Exam tip: connect the target total with the sum formula.

The production is the decreasing AP (240,225,210,\ldots), and (S_{12}=1890). Exam tip: in a decreasing AP, (d) is negative.

The page numbers are (30,40,50,\ldots), and (S_n=750) gives (n=10). Exam tip: set the reading target equal to (S_n).

The coin numbers are (50,75,100,\ldots), and (S_{12}=2250). Exam tip: treat level rewards as AP terms.

The amount filled each minute forms an AP with first term \(a=25\) and common difference \(d=8\). In the \(n\)th minute, the amount is \(a_n=25+(n-1)\times 8\). Solving \(25+(n-1)\times8=121\) gives \(n-1=12\), so \(n=13\). Hence, 121 litres are filled in the 13th minute. In the 12th minute, the amount would be only 113 litres. Exam tip: for a specified term of an AP, use \(a_n=a+(n-1)d\).

The instalments are (5000,5500,6000,\ldots), and (S_6=37500). Exam tip: treat instalments as AP terms.

The flower numbers are (18,24,30,\ldots), and (S_{11}=528). Exam tip: take the number of layers as the number of terms.

The page numbers are (15,20,25,\ldots), and (S_9=315). Exam tip: use the sum formula for total work.

The expenses are (800,1000,1200,\ldots), and (S_7=9800). Exam tip: add daily expenses as AP terms.

The sales form the decreasing AP (200,190,180,\ldots), and (S_{15}=1950). Exam tip: keep (d) negative for decreasing sales.

In option B, consecutive differences are equal: \(18-12=6\) and \(24-18=6\), so it is an AP. A and C involve multiplicative growth, while D has differences 3 and 5. In exams, check equal differences first.

Using (S_{12}=3720) and (d=40) gives (a=90). Exam tip: find the first term from total and difference.

Putting (S_n=840) in the sum formula gives (n=15). Exam tip: choose the correct positive integer answer.

The governing concept is the sum of the first n terms of an arithmetic progression. The gaps are 18, 20, 22, and so on; hence a = 18, d = 2 and n = 15. The fifteenth gap is a_15 = 18 + (15 - 1)2 = 46 metres. Using S_n = n/2 [2a + (n - 1)d], we obtain S_15 = 15/2 [36 + 14 × 2] = 15/2 × 64 = 15 × 32 = 480 metres. Therefore option C is correct. The same result follows from 15 × (first gap + last gap)/2 = 15 × (18 + 46)/2. The other choices do not satisfy the AP sum.

The member numbers are (15,20,25,\ldots), and (S_n=585) gives (n=13). Exam tip: find the number of months from the target total.

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

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

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

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

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