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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 (see Fig). In how many rows are the 200 logs placed and how many logs are in the top row?
Asked by Ravi Last Modified
Answer 
the number of logs stacked row wise form an arthimetic progression:
20,19,18,...
a=20
d=t2-t1=19-20= -1
Sn=(n/2)*(2a+(n-1)d)
200=(n/2)*(2*20+(n-1)(-1))
200=(n/2)(40-n+1)
400=n(41-n)
400=41n-n*n
n*n-41n+400=0
n*n-16n-25n+400=0
n(n-16)-25(n-16)=0
(n-16)(n-25)=0
n=16 or n=25
tn=a+(n-1)d
t16=20+(16-1)(-1)
t16=5
Checking for n=25
t25=20+(25-1)(-1) = -5
Seq is
20,19,18,....5,4,3,2,1,0,1,2,3,4
This is not applicable
Answer
16 rows
5 logs on top.
read less
Sherine Das
Well Qualified and Certified Mathematics B.Ed(M.G Usty)Teacher with more than13 years of Experience
Logs are stacked in the decreasing order of numbers,which are in arithmetic progression with first term(a)=20;Common Difference=19-20=-1 and Sum to n-terms(Sn)=200;n=number of rows in which logs are placed.
∴ Sn =n/2[2a+(n-1)d] ⇒ 200 =n/2[40+(n-1)∗-1]
400=n[40+1-n] =41 n- n2
⇒ n2-41n+400=n2-16n-25n+400 [splitting the middle term for solving the quadratic equation]
=n(n-16)-25(n-16) =( n-16)(n-25) =0 ⇒n-16=0 or n-25=0.
⇒ n=16 or n=25.
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