Question

200 logs are stacked in the following manner:20 logs in the bottom row,19 in the next row,18 in the next row to it and so on.In how many rows are the 200 logs placed and how many logs are in the top row?

Answer 1

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Answer 2

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.