ABC is a right angle triangle at B and BC - 10- If 100 points L1- L2- L3- - L100 on AB are such that AB is divided into 101 equal parts and L1M1- L2M2- - L100M100 are line segments parallel to BC and points M1- M2- M3- - -M100 are on AC- then the sum of L1M1- L2M2- - L100M100 is

Using similar triangle properties, AL_1AB = L_1M_1BC ⇒1101 = L_1M_110 ⇒ L_1M_1 nbsp; = 10101 Similarly, AL_2AB = L_2M_2BC ⇒ nbsp;L_2M_2 = nbsp; 2 ×10101 ...

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Byju's Answer

Standard XII

Mathematics

Sum of n Terms

ABC is a righ...

Question

$A B C$ is a right angle triangle at $B$ and $B C = 10.$ If $100$ points $L_{1}, L_{2}, L_{3}, \dots, L_{100}$ on $A B$ are such that $A B$ is divided into $101$ equal parts and $L_{1} M_{1}, L_{2} M_{2}, \dots, L_{100} M_{100}$ are line segments parallel to $B C$ and points $M_{1}, M_{2}, M_{3}, \dots, M_{100}$ are on $A C,$ then the sum of $L_{1} M_{1}, L_{2} M_{2}, \dots, L_{100} M_{100}$ is

500

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600

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350

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460

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Solution

The correct option is A $500$

Using similar triangle properties,
$\frac{A L_{1}}{A B} = \frac{L_{1} M_{1}}{B C}$
$\Rightarrow \frac{1}{101} = \frac{L_{1} M_{1}}{10}$
$\Rightarrow L_{1} M_{1} = \frac{10}{101}$

Similarly,
$\frac{A L_{2}}{A B} = \frac{L_{2} M_{2}}{B C}$
$\Rightarrow L_{2} M_{2} = 2 \times \frac{10}{101}$
and $L_{3} M_{3} = 3 \times \frac{10}{101}$

So, the required sum will be,
$S = L_{1} M_{1} + L_{2} M_{2} + L_{3} M_{3} + \dots + L_{100} M_{100}$
$\Rightarrow S = 1 \times \frac{10}{101} + 2 \times \frac{10}{101} + 3 \times \frac{10}{101} + \dots + 100 \times \frac{10}{101}$
$\Rightarrow S = \frac{10}{101} (1 + 2 + 3 + \dots + 100)$
$\Rightarrow S = \frac{10}{101} (\frac{100 \times 101}{2}) = 500$

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ABC is a right angle triangle at B and BC=10. If 100 points L1,L2,L3,…,L100 on AB are such that AB is divided into 101 equal parts and L1M1,L2M2,…,L100M100 are line segments parallel to BC and points M1,M2,M3,…,M100 are on AC, then the sum of L1M1,L2M2,…,L100M100 is