Question

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SOLVE 8th, 9th, 10th question properly and briefly.
PLEASE GIVE CONTENT QUALITY ANSWER. INCOMPLETE ANSWERS OR SPAM ANSWER WILL BE DELETED

Answer 1

8)
nth term of an A.P. = a + (n-1)d with a as first term and d as common difference. 

    m{a + (m-1)d} = n{a+ (n-1)d}
=>ma + m(m-1)d = na + n(n-1)d
=>(m-n)a = {n(n-1) - m(m-1)}d
                = (n² - n - m² + m)d
                = {(n² - m²) + (m-n)}d
                = {(n+m)(n-m) +(m-n)}d
                = (m-n)( -m-n + 1)d
=> a = (1-m-n)d

(m+n)th term = a + (m+n-1)d
                      = a - (1-m-n)d            
                      = a - a = 0

9)
Let the number be x. Then x + 12 = 160/x. Multiply x to each term. We get x² + 12 x = 160 or x² + 12 x - 160 = 0.

Or x² - 20 x + 12 x - 160 = 0
Or x(x - 20) + 12(x - 20) = 0
Or (x+12)(x-20) = 0
Or x = -12,20
Note that natural numbers are always positive. So the only correct answer is 20.

10) Extend AB and CD to meet at X. Since tangents from the same external point to a circle are equal, AX = CX and BX = DX. So AX - BX = CX - DX
=> AB = CD

Answer 2

Answer:

nth term of an A.P. = a + (n-1)d with a as first term and d as common difference.

m{a + (m-1)d} = n{a+ (n-1)d}

=>ma + m(m-1)d = na + n(n-1)d

=>(m-n)a = {n(n-1) - m(m-1)}d

= (n² - n - m² + m)d

= {(n² - m²) + (m-n)}d

= {(n+m)(n-m) +(m-n)}d

= (m-n)( -m-n + 1)d

=> a = (1-m-n)d

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

= a - (1-m-n)d

= a - a = 0

9)

Let the number be x. Then x + 12 = 160/x. Multiply x to each term. We get x² + 12 x = 160 or x² + 12 x - 160 = 0.

Or x² - 20 x + 12 x - 160 = 0

Or x(x - 20) + 12(x - 20) = 0

Or (x+12)(x-20) = 0

Or x = -12,20

Note that natural numbers are always positive. So the only correct answer is 20.

10) Extend AB and CD to meet at X. Since tangents from the same external point to a circle are equal, AX = CX and BX = DX. So AX - BX = CX - DX

=> AB = CD