The way to solve the problem of arithmetic progressions of exercise 5.3 of class 10th math, How i solve this question by simple way 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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# 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? Q.19

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We can see that the numbers of logs in rows are in the form of an A.P.20, 19, 18â€¦

For the given A.P.,

First term, aÂ = 20 and common difference,dÂ =Âa_{2}âˆ’a_{1}Â = 19âˆ’20 = âˆ’1Let a total of 200 logs be placed inÂ

nÂ rows.Thus, SÂ = 200_{n}By the sum of nth term formula,

SÂ =Â_{n}n/2Â [2aÂ +(nÂ -1)d]S= 12/2Â [2(20)+(_{12}ÂnÂ -1)(-1)]400 =Â

nÂ (40âˆ’n+1)400 =Â

nÂ(41-n)400 = 41

nâˆ’n^{2}n^{2}âˆ’41nÂ+ 400 = 0n^{2}âˆ’16nâˆ’25n+400 = 0n(nÂ âˆ’16)âˆ’25(nÂ âˆ’16) = 0(

nÂâˆ’16)(nÂ âˆ’25) = 0Either (

nÂ âˆ’16) = 0 orÂnâˆ’25 = 0nÂ = 16 orÂnÂ = 25By the nth term formula,

aÂ =Â_{n}a+(nâˆ’1)da_{16}Â = 20+(16âˆ’1)(âˆ’1)a_{16}Â = 20âˆ’15a_{16}Â = 5Similarly, the 25

^{th}term could be written as;a_{25}Â = 20+(25âˆ’1)(âˆ’1)a_{25}Â = 20âˆ’24= âˆ’4It can be seen, the number of logs in 16

^{th}Â row is 5 as the numbers cannot be negative.Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16

^{th}Â row is 5.