Word problems based on APs

This is an arithmetic-progression word problem because the number of seats increases by a constant amount, 7, from one row to the next. Here the first term is a = 34, the common difference is d = 7, and the number of terms is n = 21. The required total is S_n = n/2[2a + (n - 1)d]. Thus S_21 = 21/2[2(34) + 20(7)] = 21/2(68 + 140) = 21/2 × 208 = 2184. Hence option C is correct. Options A, B and D result from incorrect substitution or arithmetic.

The amounts are the AP (90,120,150,\ldots) and (S_{14}=3990). Exam tip: treat the number of days as the number of terms.

The packet numbers form a decreasing AP and (S_{14}=4725). Exam tip: take a decrease as a negative common difference.

Using (S_{16}=1320) and (a=45) gives (d=5). Exam tip: find the common difference from total and first term.

The numbers of flowers form an AP with first term \(a=42\) and common difference \(d=11\). The number of flowers in the \(n\)th layer is \(a_n=42+(n-1)\times 11\). Putting \(a_n=207\), \(42+(n-1)\times 11=207\), so \((n-1)\times 11=165\), giving \(n-1=15\) and \(n=16\). Hence, the 16th layer has 207 flowers. The 15th layer would have only \(196\) flowers. Exam tip: when a position or layer is asked, equate the target value to \(a_n\) and solve for \(n\).

The production is the AP (70,83,96,\ldots) and (S_{12}=1698). Exam tip: treat each hour's production as one term.

Using (S_{18}=1485) and (d=5) gives (a=40). Exam tip: solve the sum formula backward to find the first term.

The amount of water forms a decreasing AP with \(a=210\) and \(d=-12\). On the \(n\)th day, \(a_n=210+(n-1)(-12)\). Solving \(210-12(n-1)=66\) gives \(n-1=12\), so \(n=13\). Therefore, 66 litres remain on the 13th day. On the 12th day, the amount is 78 litres, so option 12 is not correct. Exam tip: use a negative common difference \(d\) for a decreasing AP.

The fares form the AP (85,115,145,\ldots) and (S_{11}=2585). Exam tip: treat each kilometre's fare as one term.

Using (S_{20}=1230) and (a=14) gives (d=5). Exam tip: daily increase can be found from total practice.

The coin numbers are (75,110,145,\ldots) and (S_{16}=5400). Exam tip: treat level rewards as an AP.

The weekly sales form an AP with first term \(a=95\) and common difference \(d=18\). The sale in the \(n\)th week is \(a_n=95+(n-1)18\). Setting this equal to 311 gives \(95+(n-1)18=311\), so \(18(n-1)=216\), \(n-1=12\), and \(n=13\). Therefore, 311 kg will be sold in the 13th week. In the 12th week, the sale would be \(293\) kg, so it is a close but incorrect option. Exam tip: equate the target quantity to \(a_n\) and solve for \(n\).

The lights form the decreasing AP (96,92,88,\ldots) and (S_{18}=1116). Exam tip: in decreasing rows the common difference is negative.

The distances form the AP (28,35,42,\ldots) and (S_{15}=1155). Exam tip: treat each hour's distance as one term.

Using (S_{14}=2590) and (d=12) gives (a=107). Exam tip: find the first day from total and increase.

The amounts are (60,80,100,\ldots) and (S_n=2040) gives (n=12). Exam tip: put the target total in the sum formula.

The salaries are (18000,20200,22400,\ldots) and (S_7=172200). Exam tip: add salary increases like an AP.

The box numbers are (320,300,280,\ldots) and (S_{11}=2420). Exam tip: take (d) negative for decreasing stock.

The points are (120,165,210,\ldots) and (a_{18}=885). Exam tip: take the level number as the term number.

The row sizes form an arithmetic progression. The first term is a = 25, the number of rows is n = 22, and the total is S_22 = 1243. Using S_n = n/2[2a + (n - 1)d], we get 1243 = 22/2[2(25) + 21d] = 11(50 + 21d). Dividing by 11 gives 113 = 50 + 21d, so 21d = 63 and d = 3. Therefore, option B is correct. The other choices do not satisfy the given total: substituting any of them into the sum formula produces a total different from 1243.