ABCD is a square of length a- a - N - a -1- Let L 1- L 2- L 3- - be points on BC such that BL 1- L 1 L 2- L 2 L 3-1 andM 1- M 2- M 3- - are points on CD such that CM 1- M 1 M 2- M 2 M 3-1- Then - n -1 a -1 AL n 2- L n M n 2 is equal toA- 1-2 aa-1B- 1-2 a -12 a -14 a -1C- 1-2 a a -14 a -1D- 1-2 a -122 a -1

The correct option is C 12 a (a 1)(4a 1) (AL_1)^2 +(L_1 M_1)^2 = [a^2 +1^2]+[(a 1)^2 +1^2] (AL_2)^2 +(L_2 M_2)^2 = [a^2 +2^2]+[(a 2)^2 +2^2] (AL_a 1)^ ...

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Byju's Answer

Standard XI

Mathematics

Intersection

ABCD is a squ...

Question

$A B C D$ is a square of length $a, a \in N, a > 1.$ Let $L_{1}, L_{2}, L_{3}, \dots$ be points on $B C$ such that $B L_{1} = L_{1} L_{2} = L_{2} L_{3} = \dots = 1$ and $M_{1}, M_{2}, M_{3}, \dots$ are points on $C D$ such that $C M_{1} = M_{1} M_{2} = M_{2} M_{3} = \dots = 1.$ Then $a - 1 \sum n = 1 (A {L_{n}}^{2} + L_{n} {M_{n}}^{2})$ is equal to

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

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\frac{1}{2} (a - 1) (2 a - 1) (4 a - 1)

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\frac{1}{2} a (a - 1) (4 a - 1)

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

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Solution

The correct option is C $\frac{1}{2} a (a - 1) (4 a - 1)$

$(A L_{1})^{2} + (L_{1} M_{1})^{2} = [a^{2} + 1^{2}] + [(a - 1)^{2} + 1^{2}] (A L_{2})^{2} + (L_{2} M_{2})^{2} = [a^{2} + 2^{2}] + [(a - 2)^{2} + 2^{2}] (A L_{a - 1})^{2} + (L_{a - 1} M_{a - 1})^{2} = [a^{2} + (a - 1)^{2}] + [(1)^{2} + (a - 1)^{2}]$

Therefore the required sum will be,
$= (a - 1) a^{2} + [1^{2} + 2^{2} + 3^{2} + \dots + (a - 1)^{2}] + 2 [1^{2} + 2^{2} + 3^{2} + \dots + (a - 1)^{2}] = (a - 1) a^{2} + 3 [1^{2} + 2^{2} + 3^{2} + \dots + (a - 1)^{2}] = (a - 1) a^{2} + 3 \times \frac{(a - 1) (a) (2 a - 1)}{6} = \frac{1}{2} a (a - 1) (4 a - 1)$

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Similar questions

Q. ABCD is a square of length a,

a ϵ N

, a>1. Let

L_{1}

L_{2}

L_{3}

,... be points on BC such that

B L_{1} = L_{1} L_{2} = L_{2} L_{3} = . . . = 1

and

M_{1}

M_{2}

M_{3}

,... be points on CD such that

C M_{1} = M_{1} M_{2} = M_{2} M_{3} = . . . = 1

. Then

\sum_{n = 1}^{a - 1} (A {L_{n}}^{2} + L_{n} {M_{n}}^{2})

is equal to

A B C D

is a square of length

a, a \in N, a > 1

. Let

L_{1}, L_{2}, L_{3} . . . . . .,

be point on BC such that

B L_{1} = L_{1} L_{2} = L_{2} L_{3} = . . . . . = 1

and

M_{1}, M_{2}, M_{3} . . . . . . . . .

be points on CD such that

C M_{1} = M_{1} M_{2} = . . . . . . . . . . . = 1

. Then

a - 1 \sum n = 1 (A L_{2}^{2} + L_{n} M_{n}^{2})

is equal to

Q. PQRS is a square of length a, a natural number >1. Let

L_{1}, L_{2}, L_{3}, L_{4} \dots

be points on QR such that Q

L_{1} = L_{1} L_{2} = L_{2} L_{3} = L_{3} L_{4} \dots . = 1

and

M_{1}, M_{2}, M_{3}

be points on RS such that R

M_{1} = M_{1} M_{2} = M_{2} M_{3} . . . .

. . .=1. Then,

_{n = 1}^{a - 1} \sum (P L_{n}^{2} + L_{n} M_{n}^{2})

is equal to?

Q. Factorize:

3 a^{2} - 1 - 2 a

Q. Let

a, b, c

be such that

b (a + c) \neq 0

∣ ∣ ∣ ∣ \begin{matrix} a & a + 1 & a - 1 - b & b + 1 & b - 1 c & c - 1 & c + 1 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ ∣ \begin{matrix} a + 1 & b + 1 & c - 1 a - 1 & b - 1 & c + 1 (- 1)^{n + 2} a & (- 1)^{n + 1} b & (- 1)^{n} c \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

, then the value of

n

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ABCD is a square of length a,a∈N,a>1. Let L1,L2,L3,⋯ be points on BC such that BL1=L1L2=L2L3=⋯=1 and M1,M2,M3,⋯ are points on CD such that CM1=M1M2=M2M3=⋯=1. Then a−1∑n=1(ALn2+LnMn2) is equal to