Question

200 logs are stacked in the following manner that is 20 locks in the bottom row, 19 in the next row, 18 in the room next to it and so on in how many rows are the 200 blocks placed and how many logs are in the top floor?

Answer 1

here's ur answer
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Answer 2

Answer: There are 5 logs in the top floor.

Step-by-step explanation:

Since we have given that

20 logs in the bottom row,

19 logs in the next row,

18 in the next and so on.

so, it becomes an A.P.,

20,19,18..........

Total logs = N= 200

So, we will use formula for "nth term":

$S_n=\dfrac{n}{2}(2a+(n-1)d)\\\\200=\dfrac{n}{2}(2\times 20+(n-1)\times -1)\\\\400=n(40-n+1)\\\\400=40n-n^2+n\\\\n^2-41n+400=0\\\\n^2-25n-16n+400=0\\\\n(n-25)-(n-16)=0\\\\n=16,25$

But , $a_{25}=20+(25-1)\times -1=20-24=-4$

which is not possible.

so, number of logs must be equal to 16.

$a_{16}=20+(16-1)\times -1\\\\a_{16}=20-15\\\\a_{16}=5$

Hence, there are 5 logs in the top floor.