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 ( as shown in the figure given below ) .in how many rows are the 200 logs placed and how many logs are there in the top row?
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Answered by
0
Use the formula Sn = n/2[2a+(n-1)d]
Sn is 200,
d is -1,
a is t1 = 20.
You will end up with an algebraic equation.
Solving it will give the required answer.
Also we can then we can can calculate the logs in the top row. The solution is big so I haven't written it directly... sorry for the inconvenience.
Sn is 200,
d is -1,
a is t1 = 20.
You will end up with an algebraic equation.
Solving it will give the required answer.
Also we can then we can can calculate the logs in the top row. The solution is big so I haven't written it directly... sorry for the inconvenience.
Answered by
1
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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