200 logs are stacked in the following manner 20 loss in the bottom row 19 in the next 18 in row above fit and so on how many Rose are the 200 locks place and how many locks are there in the top row
Answers
Answer:
First row = a1 = 20 logs
Common difference = a2 - a1 = 19 - 20 = -1
Find the nth term:
an = a1 + (n - 1)d
an = 20 + (n - 1)(-1)
an = 20 - n + 1
an = 21 - n
Find nth:
Sn = 200
Sn = n/2 (a1 + an)
200 = n/2 (20 + 21 - n)
400 = n(41 - n)
400 = 41n - n²
n² - 41n + 400 = 0
(n - 16)(n - 25) = 0
n = 16 or n = 25
When n = 16
a16 = 21 - 16
a16 = 5
when n = 25
an = 21 - 25 = -4 (rejected, an cannot be negative)
Answer: There are 16 rows and the top row has 5
Answer:
There are 5 logs in the top row
Explanation:
There are 20 logs in the first row [ bottom] ,
second row = 19 and so on...
It forms an AP
Where a=20 and d= -1
Sum of n terms of an AP = 200
Sn= n/2 [2a + {n-1}d]
=n/2 [2 x 20 + {n-1}{-1}]=200
=n/2 [40-n+1]=200
=n[41-n]=400 {∵200 x 2 = 400 }
=41n-n²=400
=n²-41n + 400=0
=[n - 25][n + 16]= 0
=n=25 or 16
If n=25 , then the number of logs in the 25th row is equal to 25th term of AP with first term = 20 and d= -1
∴Then the number of logs in the 25th row =
a + 24d
=20 - 24
= -4
So it is not possible
Therefore, n=16
Logs are placed in 16 rows
Number of logs in top row = Number of logs in 16th row
=16th term of an AP with a = 20 and d = -1
=a + 15d
20 + 15[ -1 ]
=20 - 15
=5
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