1.

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

Answer»

`3/13`
`5/13`
`7/13`
`9/13`

Solution :`sum_(i=1)^(N-1)(x_(i)-2isqrt(x_(i)-i^(2)))=0`
`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`


Discussion

No Comment Found

Related InterviewSolutions