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New answer posted

8 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

T r + 1 = 1 2 0 C r 4 1 2 0 r 4 . 5 r / 6

For to be rational

r should be multiple of 6

Number of such terms = 21

New answer posted

8 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Bowlers              Batsmen             Wicket Keepers

                   (6)                        (7)                               (2)

     

...more

New answer posted

8 months ago

0 Follower 7 Views

R
Raj Pandey

Contributor-Level 9

A r e a = 1 2 ( 5 1 ) 9 5 4 1 5 1 2 5 x 2 d x

= 1 8 ( 1 4 + 1 0 5 ) 1 2 c o s 1 1 5 0 ( s i n 2 θ ) d θ

A = 1 4 8 + 5 4 5 ( 5 4 c o s 1 1 5 1 2 )

= 5 4 5 5 4 5 4 c o s 1 1 5

α = 5 4 , β = 5 4 , γ = 5 4

| α + β + γ | = 5 4

New answer posted

8 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

| a | = | b | = | c | = l

a . b = b . c = c . a = 0

| a + b + c | 2 = 3 l 2

l 2 = 3 l 2 c o s θ c o s θ = 1 3

36 cos2 2q = 36 ( 2 3 1 ) 2 = 4

New answer posted

8 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

R 1 R 1 R 2 , R 2 R 2 R 3

Δ = | a c + 1 b c a b b d 1 c d b c x b + d x + d x + c |

C 1 C 1 C 2 & C 2 C 2 C 3

Δ = | a + 1 b 2 b c a a b b 1 c 2 c b d b c b d c x + c | = | λ + 1 0 λ λ 1 0 λ b λ x + c | , R 1 R 1 R 2

= | 2 0 0 λ 1 0 λ b λ x + c | = 2 λ 2 = 2 λ 2 = 1

New answer posted

8 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

0 = -16 m + c . (i)

| m ( 1 0 ) + C | m 2 + 1 = 2 . . . . . . . . . . . . . ( i i )

(i) & (ii) Þ 36 m2 = 4 (m2 + 1)

m = 1 2 2 , C = 8 2

4 2 ( m + c ) = 3 4

 

New answer posted

8 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

0 = 16 m + c . (i)

  | m ( ? 1 0 ) + C | m 2 + 1 = 2 . . . . . . . . . . . . . ( i i )

(i) & (ii) 36 m2 = 4 (m2 + 1)

m = 1 2 2 , C = 8 2

4 2 ( m + c ) = 3 4

New question posted

8 months ago

0 Follower 2 Views

New answer posted

8 months ago

0 Follower 10 Views

R
Raj Pandey

Contributor-Level 9

A = [ 1 1 0 0 1 1 0 0 1 ]

B = 7A20 – 20A7 + 2l

A 2 [ 1 1 0 0 1 1 0 0 1 ] [ 1 1 0 0 1 1 0 0 1 ] = [ 1 2 1 0 1 2 0 0 1 ]

A 3 = [ 1 2 1 0 1 2 0 0 1 ] [ 1 1 0 0 1 1 0 0 1 ] = [ 1 3 3 0 1 3 0 0 1 ]

A 4 = [ 1 3 3 0 1 3 0 0 1 ] [ 1 1 0 0 1 1 0 0 1 ] = [ 1 4 6 0 1 4 0 0 1 ]

a 1 3 o f A , A 2 , A 3 , . . . . . . .

are 0, 1, 3, 6,.

For 0, 1, 3, 6, .

S n = 0 + 1 + 3 + 6 + . . . . + t n

S n = 0 + + 3 + . . . . + t n 1 + t n

t n = 1 + 2 + 3 + . . . + ( t n t n 1 )

= ( n 1 ) . n 2

b 1 3 = 7 ( 1 9 0 ) 2 0 ( 2 1 ) + 0

= 1330 – 420 = 910

 

 

 

 

New question posted

8 months ago

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