ABC is a right angle triangle at B and BC -10- If 100 points L 1- L 2- L 3- - L 100 on AB are such that AB is divided into 101 equal parts and L 1 M 1- L 2 M 2- - L 100 M 100 are line segment parallel to BC and points M 1- M 2- M 3- - M 100 are on AC- then the sum ofL 1 M 1- L 2 M 2- - L 100 M 100 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 segment 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

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Solution

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

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 segment parallel to BC and points M1,M2,M3,⋯M100 are on AC, then the sum of L1M1,L2M2,⋯,L100M100 is

Using similar triangle properties, AL1AB=L1M1BC⇒1101=L1M110⇒L1M1=10101 Similarly, AL2AB=L2M2BC⇒L2M2=2×10101⇒L3M3=3×10101 So, the required sum will be, S=L1M1+L2M2+L3M3+⋯+L100M100⇒S=1×10101+2×10101+3×10101+⋯+100×10101⇒S=10101(1+2+3+⋯+100)⇒S=10101(100×1012)=500