200 logs start in such a way that that 20 log in the bottom row 19 Log in next row 18 log in the next row and so on. how many rows are formed and how many logs are there in
the top row
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Answered by
1
A.P=20,19,18....
a=20 d=-1 sn=200 n=?
sn= n/2 [2a+(n-1)d]
200=n/2[2×20+(n-1)(-1)]
200=n/2[41-n]
400=41n-n^2
n^2-41n+400=0
n^2-25n-16n+400=0
n(n-25)-16(n-25)=0
n=25 or n= 16
n will be 16 bcoz the ap is in decreasing order This implies there r 16 rows and 5 logs placed in top row
a=20 d=-1 sn=200 n=?
sn= n/2 [2a+(n-1)d]
200=n/2[2×20+(n-1)(-1)]
200=n/2[41-n]
400=41n-n^2
n^2-41n+400=0
n^2-25n-16n+400=0
n(n-25)-16(n-25)=0
n=25 or n= 16
n will be 16 bcoz the ap is in decreasing order This implies there r 16 rows and 5 logs placed in top row
Answered by
0
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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