Word problems based on APs

The number of drawings is (16,24,32,\ldots) and (a_{15}=128). Exam tip: work done on the (n)th day is the (n)th term.

The marks form the AP (5,11,17,\ldots) and (S_{17}=901). Exam tip: take the number of questions as (n) and find the sum.

The light numbers are (60,75,90,\ldots) and (S_8=900). Exam tip: treat each gate as one term.

The instrument numbers are (25,35,45,\ldots) and (S_n=960) gives (n=12). Exam tip: when a total target is given, find the number of terms.

The distances form the AP (20,26,32,\ldots) and (S_{16}=1040). Exam tip: treat each hour's distance as a term.

The weekly sales are (85,100,115,\ldots) and (S_{14}=2555). Exam tip: use the sum formula to add sales across weeks.

The governing concept is the sum of an arithmetic progression. Daily incomes are 700, 850, 1000, and so on, with first term a = 700, common difference d = 150, and n = 8. The total is S₈ = n/2[2a + (n−1)d]. Therefore S₈ = 8/2[2(700) + 7(150)] = 4[1400 + 1050] = 4 × 2450 = 9800 rupees. Thus option B is correct. The question asks for total income, not the income on the eighth day; using only the eighth-day amount would miss the required addition of all eight daily incomes. The other choices result from arithmetic errors.

This problem uses the nth-term formula for a decreasing arithmetic progression. The first term is a = 180 and the common difference is d = −15 because every higher floor has 15 fewer windows. Set the nth term equal to 45: 45 = 180 + (n−1)(−15). This gives −135 = −15(n−1), so n−1 = 9 and n = 10. The sequence confirms the result: the first floor has 180 windows, the second 165, and after nine decreases the tenth floor has 45. Hence option C is correct. Options A, B and D give 75, 60 and 30 windows respectively.

The page numbers are (30,38,46,\ldots) and (S_{16}=1440). Exam tip: use the sum formula for total printing.

The cloth sales form (100,125,150,\ldots) and (S_9=1800). Exam tip: use the sum formula for sales across all days.

The member numbers are (18,27,36,\ldots) and (S_{20}=2070). Exam tip: add all days for total members.

Using (S_{16}=1120) and (a=25) gives (d=6). Exam tip: find the common difference from total and first term.

The plate numbers are (20,26,32,\ldots) and (S_{15}=930). Exam tip: treat each table as one term of the AP.

The doses form the AP (10,14,18,\ldots) and (S_{18}=792). Exam tip: add the terms for the total dose.

The water amounts are (120,150,180,\ldots) and (S_{10}=2550). Exam tip: treat hourly filled water as AP terms.

This is an arithmetic progression with first term 110 and common difference 15. The sales on the nth day are \(a_n=110+(n-1)\times15\). Putting \(a_n=320\), we get \(110+15(n-1)=320\), so \(15(n-1)=210\), \(n-1=14\), and \(n=15\). Therefore, 320 newspapers will be sold on the 15th day. The nearby option 13 is incorrect because sales on that day would be \(110+12\times15=290\). Exam tip: In day-number AP questions, remember to use \(n-1\).

Putting (S_n=2328) in the sum formula gives (n=24). Exam tip: when total seats are given, form an equation for (n).

The salaries form the AP (35000,36800,38600,\ldots), and (S_9=379800). Exam tip: add all monthly salaries using the sum formula.

The production is the decreasing AP (420,402,384,\ldots) and (S_{15}=4410). Exam tip: treat a decrease as a negative common difference.

For the first 20 AP terms, \(S_n=\frac{n}{2}[2a+(n-1)d]\). Here \(n-1=19\), so \(S_{20}=10[50+57]=1070\). Twenty rows create only 19 increases. Exam tip: check the \((n-1)d\) term carefully.