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Ex 5.3, 19 - 200 logs are stacked in following manner: - Ex 5.3

 

  1. Chapter 5 Class 10 Arithmetic Progressions
  2. Serial order wise
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Ex 5.3 ,19 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row? Number of logs in 1st row = 20 Number of logs in 2nd row = 19 Number of logs in 3rd row = 18 Hence the series is 20, 19, 18, …….. Since difference is same, it is an AP We need to find in how many rows can 200 logs be place, i.e. for Sn = 200, we need to find n We can use formula Sn = 𝑛/2 (2𝑎+(𝑛−1)𝑑) Here, Sn = 200, a = 20 , d = – 1 We need to find n Putting values Sn = 𝑛/2 (2𝑎+(𝑛−1)𝑑) 200 = 𝑛/2 (2×20+(𝑛−1)×−1) 200 = 𝑛/2 (40+ 𝑛(−1)−1(−1)) 200 = 𝑛/2 (40− 𝑛+1) 200 = 𝑛/2 (41−𝑛) 200 ×2=𝑛(41−𝑛) n2 – 41n + 400 = 0 We factorize by splitting the middle term n2 – 16n – 25n + 400 = 0 n (n – 16) – 25 (n – 16) = 0 (n – 25) (n – 16) = 0 Hence, n = 16 or 25 Note:- There are 20 logs in 1 row 19 logs in 2 row 18 logs in 3 row Going on there will 5 logs in 16 row & there will be – 4 logs in 25 row Since – 4 logs is not possible So, n = 25 is not possible solution This can be checked by using formula an = a + (n – 1) d n = 25, a = 20, d = – 1 Putting values a25 = 20 + (25 – 1) ×−1 a25 = 20 + (24) ×−1 a25 = 20 – (24) a25 = – 4 Since the number of logs cannot be negative, n = 25 is not a possible solution Thus, n = 16 is the only solution So, there are 16 rows of logs, & top row has 5 logs

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