Question

If m,n and t are in

a.p. then prove that (m+2n-t)(2n+t)(2n+t-m)(t+m-n)=4mnt

Answer 1

you did mistake in typing .
question is ----> If m,n and t are in a.p. then prove that (m+2n-t)(2n+t-m)(t+m-n)=4mnt

solution :- m, n and t are in AP
so, common difference will be constant .
n - m = t - n
t + m = 2n
t + m - n = 2n - n = n
(t + m - n ) = n ------------(1)

t + m = 2n
t + m + m = 2n + m
t + 2m - t = m + 2n- t
2m = ( m + 2n - t )
( m + 2n - t) = 2m ----------(2)

t + m = 2n
t + m + t = 2n + t
2t + m - m = 2n + t - m
(2n + t - m ) = 2t ----------(3)

Now, LHS = (m + 2n - t )(2n + t - m )(t + m - n )
putting equations (1), (2) and (3)
= 2m × 2t × n
= 4mnt = RHS
Hence proved//

Answer 2

Step-by-step explanation:

solution :- m, n and t are in AP

so, common difference will be constant .

n - m = t - n

t + m = 2n

t + m - n = 2n - n = n

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

t + m = 2n

t + m + m = 2n + m

t + 2m - t = m + 2n- t

2m = ( m + 2n - t )

( m + 2n - t) = 2m ----------(2)

t + m = 2n

t + m + t = 2n + t

2t + m - m = 2n + t - m

(2n + t - m ) = 2t ----------(3)

Now, LHS = (m + 2n - t )(2n + t - m )(t + m - n )

putting equations (1), (2) and (3)

= 2m × 2t × n

= 4mnt = RHS

Hence proved