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Class 10 > Learn Exercise 5.3 > 200 logs are stacked in the following manner: 20 logs...
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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?

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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...
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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
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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 ⇒ 200 =n/2 400=n =41...
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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. read less
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