ABC is a right-angled triangle in which `angleB=90^(@) and BC=a.` If n points `L_(1),L_(2),…,L_(n)` on AB is divided in n+1 equal parts and `L_(1)M_(1), L_(2)M_(2),…,L_(n)M_(n)` are line segments paralllel to BC and `M_(1), M_(2),….,M_(n)` are on AC, then the sum of the lengths of `L_(1)M_(1), L_(2)M_(2),...,L_(n)M_(n)` is

Options

A. `(a(n+1))/(2)`
B. `(a(n-1))/(2)`
C. `(an)/(2)`
D. none of these

Answer

C

Explanation

`(AL_(1))/(AB)=(L_(1)M_(1))/(BC)`
`=1/(n+1)=(L_(1)M_(1))/a`
`(AL_(2))/(AB)=(L_(2)M_(2))/(BC)`
`rArr2/(n+1)=(L_(2)M_(2))/a`
`rArrL_(2)M^(2)=(2a)/(n+1), etc`
Hence. The required sum is
`a/(n+1)+(2a)/(n+1)+(3a)/(n+1)+….+(na)/(n+1)`
`=a/(n+1)(n(n+1))/2`
`=(an)/2`