Word problems based on APs

The video numbers are (6,9,12,\ldots), and (S_{15}=405). Exam tip: take the number of weeks as the number of terms.

The capacities are (28,32,36,\ldots), and (S_{16}=928). Exam tip: add all rows for total capacity.

The call numbers are (120,135,150,\ldots), and (S_8=1380). Exam tip: treat calls per hour as AP terms.

The weekly sales form an arithmetic progression with first term 60 and common difference 10. Therefore, sales in the nth week are \(a_n=60+(n-1)\times10\). Putting \(a_n=150\), we get \(60+10(n-1)=150\), so \(n-1=9\) and \(n=10\). Hence, the farmer will sell 150 kg in the 10th week. In the 9th week, the sale would be only 140 kg. Exam tip: to find the position of a specified term, equate the target value to \(a_n\).

The admissions are (25,30,35,\ldots), and (S_{20}=1450). Exam tip: use the sum formula for total admissions.

The doses form the AP (5,7,9,\ldots), and (S_{14}=252). Exam tip: add the terms for the total dose.

The numbers of tiles form a decreasing AP with first term \(a=80\) and common difference \(d=-5\). The number of tiles in the \(n\)th row is \(a_n=80+(n-1)(-5)\). From \(80-5(n-1)=25\), we get \(n-1=11\), so \(n=12\). Hence, the 12th row has 25 tiles. The 11th row has 30 tiles, so it is a close but incorrect option. Exam tip: use a negative common difference when quantities decrease.

Using (S_{10}=7750) and (d=150) gives (a=100). Exam tip: the first term can be found from total and difference.

Putting (S_n=504) in the sum formula gives (n=12). Exam tip: you can also test the options in the sum formula.

The distances form the AP (12,15,18,\ldots), and (S_{10}=255). Exam tip: treat each hour's distance as a term.

The plate numbers are (9,13,17,\ldots), and (S_{13}=429). Exam tip: treat each table as one term of the AP.

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 correct answer is C, 315 pages. The hourly numbers form an arithmetic progression because the first term is 15 and each next term increases by 5. Thus the terms are 15, 20, 25, 30, 35, 40, 45, 50, and 55. For nine hours, use the sum formula \\(S_n=\\frac{n}{2}[2a+(n-1)d]\\). Here, \\(n=9\\), \\(a=15\\), and \\(d=5\\). Therefore, \\(S_9=\\frac{9}{2}[2(15)+8(5)]=\\frac{9}{2}(30+40)=\\frac{9}{2}\\times70=315\\). Option A, 285, is too small and results from an incorrect total. Option B, 300, does not equal the AP sum. Option C is correct because it includes all nine hourly outputs. Option D, 330, is too large and usually comes from adding an extra increase or using the wrong last term. Check also by pairing first and last terms: 15+55=70, and there are 4.5 pairs, giving 315. Memory cue: identify first term, common difference, and number of terms before using the sum formula.

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.

The direct answer is Option B: 90 rupees. The monthly savings form an AP: first term a is unknown, number of terms n=12, common difference d=40, and total S12=3720. Use S_n=n/2[2a+(n-1)d]. Substitution gives 3720=12/2[2a+11(40)]=6[2a+440]. Divide by 6: 620=2a+440. Subtract 440: 2a=180. Divide by 2: a=90. Therefore the first month’s saving is 90 rupees. Option B is correct. Option A, 80, would give a smaller total; Option C, 100, would give a larger total; Option D, 110, would make the total still larger. A quick check is useful: the last saving is 90+11(40)=530, and the average of first and last terms is (90+530)/2=310; 12×310=3720. Memory cue: total AP = number of terms × average term.

The correct answer is C, 15 rows. The seat counts form an arithmetic progression with first term \\(a=35\\), common difference \\(d=3\\), and total \\(S_n=840\\). Use \\(S_n=\\frac{n}{2}[2a+(n-1)d]\\). Substitution gives \\(840=\\frac{n}{2}[70+3(n-1)]\\), so \\(840=\\frac{n}{2}(67+3n)\\). Multiplying by 2 gives \\(1680=n(67+3n)\\), or \\(3n^2+67n-1680=0\\). Testing the positive whole-number option, \\(n=15\\) gives \\(S_{15}=\\frac{15}{2}[70+14(3)]=\\frac{15}{2}(112)=840\\). Therefore C is correct. Option A gives \\(S_{13}=663\\), so it is too small. Option B gives \\(S_{14}=749\\), also too small. Option D gives \\(S_{16}=936\\), too large. The number of rows must be a positive whole number, not a fraction. Memory cue: for an AP seating problem, write first term, increase, total, then verify the chosen row count in the sum formula.

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.