Question
21 is divided into three parts which are in arithmetic
progression (A.P.) in such a way that the sum of their square is 155. Find the smallest part.Solution
Let the three numbers be (a-x), (a), (a +x) ATQ- a-x +a+ a+ x = 21 3a = 21 a =7 So, numbers 7-x,7,7+x ATQ- (7-x) ² +49 +(7+x) ² = 155 49+x²-14x+49+49+x²+14x = 155 2x² = 155-147 =8 x² = 4 x= 2 So, find the series - 5,7,9. the smallest part of the series =5
More Divisibility rules Questions
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...5000, 4500, 3600, 2480, 1512, 756Â Â Â
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125, 150, 250, 475, 875, 1500
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... 7    12     33    126    635  Â
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