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                                    Let x_(i) epsilonR,i=1,2,3……….n are numbers such that sum_(i=1)^(n)isqrt(x_(i)-i^(2))=(sum_(i=1)^(n)x_(i))/2 and x_(1)+x_(2)+……….+x_(n)=280 Probability that a randomly selected triangle formed by vertices of a 2n+1 sided regular polygon is isosceles is | 
                            
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Answer»  `3/13` `sum_(i=1)^(n-1)(sqrt(x_(i)-i^(2)))^(2)-2isqrt(x_(i)-i^(2))+i^(2)=0` `sum_(i=1)^(n-1)(sqrt(x_(i)-i^(2))-i)^(2)=0` so, `x_(i)=2i^(2)` Now, `x_(1)^(2)+….+x_(n)^(2)=280` `2[1^(2)+2^(2)+........n^(2)]=280` `n=7` `y_(1)+y_(2)+y_(3)=7` `y_(1)^(1)+y_(2)^(1)+y_(3)^(1)=4` `.^(4+3-1)C_(3)=.^(6)C_(3)=20` Total triangles formed `=.^(15)C_(3)=(15xx14xx13)/6` `N` of isosceles triangles formed `=15xx7` probability `=(15xx7)/(15xx14xx13)xx6` `3/13`  | 
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