19, 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next ron,
18 in the row next to it and so on (see Fig. 1.5). In how many rows are the 200 logs placed
and how many logs are in the top row?
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Step-by-step explanation:
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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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