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 179. 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 = 179 49+x²-14x+49+49+x²+14x = 179 2x² = 179-147 =32 x² = 16 x= 4 So, find the series = 3,7,11. the smallest part of the series =3
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