Question

hey, does anyone know ques - 3 , 7 and 8 . please answer me its urgent

Answer 1

Answer:

Step-by-step explanation:

Answer-3 and 7

Answer 2

Answer:

Step-by-step explanation:

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3) cos A = sin A / tan A

= (m sin B) / (n tan B)

=> n cos A = m (sin B / tan B) = m cos B

n² cos² A + sin² A = m² cos² B + m² sin² B = m² ( cos² B + sin² B ) = m²

=> (n²-1) cos² A + cos² A + sin² A = m²

=> (n²-1) cos² A + 1 = m²

=> cos² A = (m²-1) / (n²-1)

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7)

Let the numbers be a, a+d, a+2d, a+3d.

Given: 4a+6d = 32 => 2a + 3d = 16 => 2a = 16 - 3d   ... (1)

Given: a(a+3d) / (a+d)(a+2d) = 7 / 15

=> 15a(a+3d) = 7(a+d)(a+2d)

=> 15a² + 45ad = 7a² + 21ad + 14d²

=> 8a² + 24ad = 14d²

=> 4a² + 12ad = 7d²

=> (2a)² + 6(2a)d = 7d²                      [ now use equation (1)... ]

=> (16-3d)² + 6d(16-3d) = 7d²

=> 256 - 96d + 9d² + 96d - 18d² = 7d²

=> 256 = 16d²

=> d² = 16

=> d = ±4

Can take d = 4 since using d = -4 will be the same numbers but in reverse order.

From (1),

a = (16-3d)/2 = (16-12)/2 = 4/2 = 2

So the numbers are:

2, 6, 10, 14

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8)

Let a be the first term and d the common difference.

Given:

m-th term = 1/n   <=>   a + (m-1)d = 1/n

n-th term = 1/m   <=>   a + (n-1)d = 1/m

Assume m ≠ n, otherwise the claim in the problem isn't even true!

Subtracting one equation above from the other gives

(m-n)d = 1/n - 1/m = (m-n) / mn

=> d = 1 / mn          [ can divide by (m-n) since m ≠ n ]

Using one of the equations above gives

a = 1/n - (m-1)d = 1/n - (m-1)/mn = 1/n - 1/n + 1/mn = 1 / mn

The sum of the first mn terms is

mna + mn(mn-1)d/2

= 1 + (mnd) (mn-1)/2

= 1 + (mn-1)/2

= (2+mn-1)/2

= (mn+1) / 2