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Example 16 - A manufacturer of TV sets produced 600 sets - Examples

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  1. Chapter 5 Class 10 Arithmetic Progressions
  2. Serial order wise
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Example 16 A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year , find: (i) the production in the 1st year Since production increases by a fixed number every year, it is an AP From (1) & (2) 600 – 2d = 700 – 6d 600 – 700 = 2d – 6d –100 = –4d (−100)/(−4) = d 25 = d d = 25 Putting the value of d in (1) a = 600 – 2d a = 600 – 2(25) a = 600 – 50 a = 550 So, production in 1st year is 550 sets Example 16 (ii) the production in the 10th year We need to find production in the 10th year i.e. a10 We know that an = a + (n – 1) d Here a = 550, d = 25 , n = 10 Putting these values in formula a10 = 550 + (10 – 1) (25) = 550 + (9) (25) = 550 + 225 = 775 Example 16 (ii) Total production in first 7 years We need to find total production in first 7 years i.e. S7 We know that Sum = 𝑛/2[2𝑎+(𝑛−1)𝑑] Here a = 550, d = 25 , n = 7 Putting these values in formula Sum = 𝑛/2[2𝑎+(𝑛−1)𝑑] = 7/2[2×550+(7−1)(25)] = 7/2[1100+(6)(25)] = 7/2[1100+150] = 7/2[1250] = (7 × 1250)/2 = 7 ×625 = 4375 ∴ Total production in first 7 years = 4375 Example 16 (iii) Total production in first 7 years We need to find total production in first 7 years i.e. S7 We know that Sum = 𝑛﷮2﷯[2𝑎+ 𝑛−1﷯𝑑] Here a = 550, d = 25 , n = 7 Putting these values in formula Sum = 𝑛﷮2﷯[2𝑎+ 𝑛−1﷯𝑑] = 7﷮2﷯[2×550+ 7−1﷯ 25﷯] = 7﷮2﷯[1100+ 6﷯ 25﷯] = 7﷮2﷯[1100+150] = 7﷮2﷯[1250] = 7 × 1250﷮2﷯ = 7 ×625 = 4375 ∴ Total production in first 7 years = 4375

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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