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

ABC is a right-angled triangle in which angleB=90^(@) and BC=a. If n p

1.	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
Answer» `(a(N+1))/(2)` `(a(n-1))/(2)` `(an)/(2)` none of these Solution :`(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`

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