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Sum of N Terms of an AP

ABC is a righ...

Question

$A B C$ is a right angled $△$ in which $∠ B = 90^{\circ}$ and $B C = a$ . If $n$ points $L_{1}, L_{2}, L_{3}, . . ., L_{n}$ on $A B$ are such that $A B$ is divided into $(n + 1)$ equal parts and $L_{1} M_{1}, L_{2} M_{2}, . . . L_{n} M_{n}$ are line segments parallel to $B C$ and points $M_{1}, M_{2}, M_{3}, . . ., M_{n}$ are on $A C$ , then the sum of the lengths of $L_{1} M_{1}, L_{2} M_{2}, . . . L_{n} M_{n}$ is :

A

\frac{a (n + 1)}{2}

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B

\frac{a (n - 1)}{2}

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C

\frac{a n}{2}

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D

none of these

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Solution

The correct option is C $\frac{a n}{2}$
We have,
$\frac{{A L}_{1}}{A B} = \frac{L_{1} M_{1}}{B C} \Rightarrow \frac{1}{n + 1} = \frac{L_{1} M_{1}}{a} \Rightarrow= L_{1} M_{1} = \frac{a}{n + 1}$
$\frac{{A L}_{2}}{A B} = \frac{L_{2} M_{2}}{B C} \Rightarrow L_{2} M_{2} = \frac{2 a}{n + 1}, . . .$ and so on
$\Rightarrow$ Required sum $= \frac{a}{n + 1} + \frac{2 a}{n + 1} + . . . + \frac{n a}{n + 1}$
$= \frac{a}{n + 1} [1 + 2 + . . . + n] = \frac{n a}{2}$

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