200 logs are stacked in such a way that 20 logs in the bottom 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?
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Answer:
In 16 rows 200 logs will be placed and there will be 5 logs 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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