Word problems based on APs

Correct answer: C. (30)

Using (S_{16}=960) and (d=4) gives (a=30). Exam tip: solve the sum formula for the unknown first term.

Correct answer: C. (1053)

The direct answer is C, 1053 books. The numbers form an arithmetic progression because the first number is 27 and the same number, 9, is added at every next stall. Thus the first term is a = 27, the common difference is d = 9, and the number of terms is n = 13. Use the sum formula: \(S_n=\frac{n}{2}[2a+(n-1)d]\). Substitution gives \(S_{13}=\frac{13}{2}[2(27)+12(9)]\). Now \(2(27)=54\), \(12(9)=108\), and \(54+108=162\). Therefore \(S_{13}=\frac{13}{2}\times162=13\times81=1053\). Option A, 1017, is too small and results from an incorrect total or term count. Option B, 1035, is also not the value of the required 13-term sum. Option C, 1053, matches the calculation and is correct. Option D, 1071, is too large, usually caused by adding one extra increase or using the wrong last term. Each stall represents one AP term, including the first stall. Exam cue: identify first term, common difference and number of terms before using the sum formula.

Correct answer: B. (600)

The direct answer is B: 600 rupees. The daily incomes form an arithmetic progression because the amount increases by the same number, 35 rupees, each day. The first term is \\(a=250\\), the common difference is \\(d=35\\), and the required term number is \\(n=11\\). Use \\(a_n=a+(n-1)d\\). Thus, \\(a_{11}=250+(11-1)\\times35=250+10\\times35=250+350=600\\). Option A, 565, would result from using only nine increases, so it is too small. Option B is correct because there are ten increases from day 1 to day 11. Option C, 635, uses one extra increase and is the income for day 12. Option D, 670, uses still more increases and is not the eleventh-day value. The common mistake is multiplying the difference by 11 instead of by 10; the first day already has the starting income. Remember: the nth day has exactly \\(n-1\\) increases after day one.

Correct answer: C. (10)

The deposits are (200,250,300,\ldots) and (S_n=4250) gives (n=10). Exam tip: form an (S_n) equation for the target amount.

Correct answer: D. (9030)

The costs form the decreasing AP (1000,955,910,\ldots) and (S_{12}=9030). Exam tip: take (d) negative for decreasing costs.

Correct answer: A. (7200)

The visits form the AP (500,575,650,\ldots) and (S_9=7200). Exam tip: add daily visits as terms.

Correct answer: D. 14th minute

The amount filled each minute forms an AP with first term \(a=36\) and common difference \(d=12\). The amount filled in the \(n\)th minute is \(a_n=36+(n-1)\times12\). Setting this equal to 192 gives \(36+(n-1)\times12=192\), so \((n-1)\times12=156\), \(n-1=13\), and \(n=14\). Hence, 192 litres are filled in the 14th minute. In the 13th minute, the amount would be only 180 litres. Exam tip: when the question asks for the amount in a particular minute, use the \(n\)th-term formula, not the sum formula.

Correct answer: B. (1920)

The dot numbers are (11,17,23,\ldots) and (S_{24}=1920). Exam tip: use the sum formula for the total of a pattern.

Correct answer: C. (11)

Using (S_{12}=990) and (a=22) gives (d=11). Exam tip: find the common difference from total and first term.

Correct answer: C. (110)

The item numbers are (500,470,440,\ldots) and (a_{14}=110). Exam tip: keep (d) negative in a decreasing AP.

Correct answer: C. (17400)

The direct answer is option C: 17,400 rupees. The daily wages form an arithmetic progression because the increase is the same each day. Here first term is a = 600, common difference is d = 80, and number of days is n = 15. Use the sum formula \\(S_n=\frac{n}{2}[2a+(n-1)d]\\). Thus \\(S_{15}=\frac{15}{2}[2(600)+14(80)]\\ =\frac{15}{2}[1200+1120]\\ =\frac{15}{2}\times2320=15\times1160=17400\\). Option A, 16,800, is too small and does not result from the correct sum. Option B, 17,100, also comes from an incorrect calculation. Option C matches the formula. Option D, 17,700, is too large and reflects another arithmetic error. The last wage is 600 + 14(80) = 1,720, which also confirms that the average wage is (600 + 1,720)/2 = 1,160 and total is 15 × 1,160 = 17,400. Memory cue: constant increase means AP; use first plus last, divided by two, multiplied by the number of terms.

Correct answer: C. (6750)

The visits form the decreasing AP (900,850,800,\ldots) and (S_{10}=6750). Exam tip: keep (d) negative for decreasing numbers.

Correct answer: D. (275)

The drawing numbers are (55,66,77,\ldots) and (a_{21}=275). Exam tip: work on the (n)th day is found by the (n)th term.

Correct answer: B. (3367)

The page numbers are (130,147,164,\ldots) and (S_{14}=3367). Exam tip: use the sum formula for total printing.

Correct answer: C. (16)

The practice times are (45,60,75,\ldots) and (S_n=2520) gives (n=16). Exam tip: set the target total equal to (S_n).

Correct answer: A. (60)

Using (S_{20}=2150) and (d=5) gives (a=60). Exam tip: find the first day from total and increase.

Correct answer: C. (3870)

The packets form the decreasing AP (300,290,280,\ldots) and (S_{18}=3870). Exam tip: keep the common difference negative for decreasing quantities.

Correct answer: B. 22

The number of seats forms an arithmetic progression with first term \(a=40\) and common difference \(d=7\). The number of seats in the \(n\)th row is \(a_n=40+(n-1)\times 7\). Setting this equal to 187 gives \(40+(n-1)\times 7=187\), so \(7(n-1)=147\), \(n-1=21\), and \(n=22\). Thus, the 22nd row has 187 seats. The 24th row would have \(40+23\times7=201\) seats, so it is not correct. Exam tip: in row-number AP questions, equate the given value to \(a_n=a+(n-1)d\).

Correct answer: C. (28)

The seats form an AP and (S_n=2520) gives (n=28). Exam tip: when total seats are given, form an (S_n) equation.

Correct answer: B. (532500)

The salaries form the AP (42000,44500,47000,\ldots) and (S_{10}=532500). Exam tip: add all monthly salaries using the sum formula.

Correct answer: D. (7434)

The production is the decreasing AP (600,578,556,\ldots) and (S_{18}=7434). Exam tip: treat a decrease as a negative common difference.

Correct answer: D. (4)

Using (S_{20}=1240) and (a=24) gives (d=4). Exam tip: find the common difference from the total and first term.

Correct answer: B. 22

This forms an arithmetic progression with first term \(a=18\) and common difference \(d=9\). The number of plants in the \(n\)th row is \(a_n=18+(n-1)\times 9\). Putting \(a_n=207\), \(18+(n-1)\times 9=207\). Thus, \((n-1)\times 9=189\), so \(n-1=21\) and \(n=22\). Hence, the 22nd row has 207 plants. The 23rd row would have 216 plants, so it is not correct. Exam tip: equate the target quantity directly to \(a_n\) and solve for \(n\).

Correct answer: C. (2760)

The daily work forms the AP (75,88,101,\ldots) and (S_{16}=2760). Exam tip: treat the daily increase as the common difference.

Correct answer: C. (11)

The donation amounts are (250,375,500,\ldots) and (S_n=9625) gives (n=11). Exam tip: form an (S_n) equation for the target total.