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 200 logs are placed and how many kids are in the top row
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SOLUTION Let the required number of rows be n. Then,
20+ 19 + 18+ .. to n terms = 200.
This is an arithmetic series in which
a = 20, d = (19-20) = -1 and S, = 200.
We know that S 12a + (n-1)d).
2
12 20 + (n − 1)(-1)) = 200
(41 - n) = 400 = n2-41n + 400 = 0
712-25n -16n + 400 = 0 => n(n-25) - 16(1-25) = 0
(n1 - 25) (11-16) = 0 = 11-25 = 0 or 11-16 = 0
n = 25 or n = 16.
Now, T = (a +24d) = 20 + 24 (-1) = -4.
This is meaningless as the number of logs cannot be negative
So, we reject the value n = 25.
11 = 16. Thus, there are 16 rows in the whole stack
Now, T = (a +150) = 20 + 15 X (-1) = 20 - 15 = 5.
Hence, there are 5 logs in the top row.
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