Sequence And Series Question 330
Question: ABC is a right angled triangle in which $ \angle B=90{}^\circ $ and $ BC=a $ . If n points $ L_1,,L_2,….L_{n} $ on AB are such that AB is divided in n + 1 equal parts and $ L_1M_1,L_2M_2,….,L_{n}M_{n} $ are line segments parallel to BC and $ M_1,,M_2,…M_{n} $ are on AC, then the sum of the lengths of $ L_1M_1,L_2M_2,….L_{n}M_{n} $ is
Options:
A) $ \frac{a(n+1)}{2} $
B) $ \frac{a(n-1)}{2} $
C) $ \frac{an}{2} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ \frac{AL_1}{AB}=\frac{L_1M_1}{BC} $
$ \therefore \frac{1}{n+1}=\frac{L_1M_1}{a} $
$ \therefore L_1M_1=\frac{a}{n+1}; $ $ \frac{AL_2}{AB}=\frac{L_2M_2}{BC} $
$ \therefore \frac{2}{n+1}=\frac{L_2M_2}{a} $
$ \therefore L_2M_2=\frac{2a}{n+1},, $ etc.
$ \therefore $ The required sum $ =\frac{a}{n+1}+\frac{2a}{n+1}+\frac{3a}{n+1}+ $ $ …+\frac{na}{n+1} $ $ =\frac{a}{n+1}(1+2+….+n)=\frac{a}{n+1}.\frac{n(n+1)}{2}=\frac{an}{2} $
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