Word problems based on APs

The units form the AP (150,175,200,\ldots) and (S_{10}=2625). Exam tip: add consumption across all months.

The number of windows forms a decreasing AP with \(a=150\) and \(d=-7\). Thus, \(a_n=150+(n-1)(-7)\). Substituting \(a_n=52\) gives \(150-7(n-1)=52\), so \(7(n-1)=98\) and \(n=15\). The 14th floor would have 59 windows and the 16th floor would have 45, so the correct answer is the 15th floor. Exam tip: write the common difference as negative for a decreasing AP.

The instalments are (7500,8750,10000,\ldots) and (S_6=63750). Exam tip: treat instalments as AP terms.

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

The book sales are (27,36,45,\ldots) and (S_{13}=1053). Exam tip: treat each stall as one term.

The income is (250,285,320,\ldots) and (a_{11}=600). Exam tip: use the (n)th term for a particular day.

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.

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

The governing idea is the nth-term formula of an arithmetic progression. The first row gives a_1 = 33, while the 18th row gives a_18 = 152. Since a_n = a + (n - 1)d, substitute the known values: 152 = 33 + (18 - 1)d = 33 + 17d. Hence 119 = 17d and d = 7. The equal increase from one row to the next is therefore 7 flags, so option C is correct. Options A, B and D would produce 18th-row values of 118, 135 and 169 respectively, so they cannot satisfy the information in the question.

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

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.

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

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

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

The successive gaps are 18, 27, 36, and so on, so they form an arithmetic progression with a = 18, d = 9, and n = 17. The question asks for the total of the gaps, not merely the distance of the 17th gap, so use S_n = n/2[2a + (n - 1)d]. Therefore S_17 = 17/2[2(18) + 16(9)] = 17/2(36 + 144) = 17/2 × 180 = 1530 metres. Thus option A is correct. The other options can arise from confusing the nth term with the sum or from an arithmetic error.

The seat counts form an arithmetic progression with first term a = 36 and common difference d = 6. If there are n rows, the total is S_n = n/2[2a + (n - 1)d]. Substitution gives 1566 = n/2[72 + 6(n - 1)] = n/2(6n + 66) = 3n(n + 11). Hence n(n + 11) = 522. Testing the positive whole-number choices, n = 18 gives 18 × 29 = 522, so the number of rows is 18. Option D is correct. The other choices do not produce the stated total when inserted into the sum formula.

The wages are (600,680,760,\ldots) and (S_{15}=17400). Exam tip: treat daily wages as an AP.

The monthly fees form an arithmetic progression because the fee rises by the same amount, 75 rupees, each month. The first term is a = 250 and the common difference is d = 75. To find the month, use the nth-term relation a_n = a + (n - 1)d and set it equal to 1075: 1075 = 250 + 75(n - 1). Thus 825 = 75(n - 1), so n - 1 = 11 and n = 12. Therefore, option B is correct. The sum formula is unnecessary because the question asks for one monthly fee, not the total fees over several months.

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

The number of instruments placed each day is an arithmetic progression. Its first term is a = 16, the number of days is n = 25, and the total is S_25 = 1600. Apply S_n = n/2[2a + (n - 1)d]: 1600 = 25/2[32 + 24d]. Multiplying by 2 gives 3200 = 25(32 + 24d); dividing by 25 gives 128 = 32 + 24d. Therefore 24d = 96 and d = 4. Option C is correct. This means each day’s placement exceeds the previous day’s placement by four instruments; the other choices do not yield a total of 1600.