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. In how many rows are the 200 logs placed and how many logs are in the top row??
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Answers
Answer:
Total Number of rows required to stack 200 logs = 16
Total number of logs in the 16th row = 5
Step-by-step explanation:
Given that the logs are stacked in rows in the following manner:
No of logs in the bottom row = a₁ = 20
Number of logs in the second row from bottom = a₂ = 19
The number of logs decrease by 1 as we go up from bottom to top, so, if we represent this relation in the form of series we get the following series:
20, 19, 18, 17..............
Now the above series is arithmetic series who common difference (d) can be found as:
d = a₂ - a₁ = 19 - 20 = -1
We need to find
In how many rows 200 logs are stacked = ?
How many logs are in top row = ?
Lets find find the nth term of the above series as show below:
= a₁ + (n - 1)d
= 20 + (n - 1)(-1)
= 20 - n + 1
= 21 - n .......(i)
Now we have to find the total number of rows when total logs become 200.
So, we can take
= 200
Using the summation formula given below:
= (a₁ + ) ...... (A)
Substituting the values of a₁ and into equation (A)
200 = (20 + 21 - n)
2x200 = n(41 - n)
400 = 41n - n²
Rearranging
n² - 41n + 400 = 0
Solving this quadratic equation by middle term breaking
n² - 16n - 25n + 400 = 0
taking commons
n(n - 16) - 25 (n - 16) = 0
(n - 16)(n - 25) = 0
n - 16 = 0 or n - 25 = 0
n = 16 or n = 25
Now, if we take total rows needed to be 25 to stack 200 logs than this negates our sequence as in each row a log must decrease from bottom to top and it is is possible only to have 20 rows only in the series. So 25 rows can not be our correct answer; therefore, our correct answer is 16 rows
So,
Total Number of rows required to stack 200 logs = 16
Now, if we plug the value of n = 16 into equation (i) we get the following expression:
= 21 - 16
= 5
So,
Total number of logs in the 16th row = 5
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.