200logs are stacked in the following manner . 20logs 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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Answer:
It is an AP such that a=20 & d=−1
S
n
=200
⇒200=
2
n
[2(20)+(n−1)(−1)]
⇒400=n[40+1−n]
⇒n
2
−41n+400=0
⇒n
2
−25n−16n+400=0
⇒(n−25)(n−16)=0
So n=16 or n=25
T
16
=a+15d=20−15=5
T
25
=a+24d=20−24=−4
Since T
25
is not possible.
∴ n=16 & 5 logs are placed in the top row.
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