Question

X+Z=2y and b^2=AC, then prove
a^y-z.b^z-x.c^x-y=1
Show me the process also guys

Answer 1

Since x + z = 2y, z = 2y - x and:
b^(z - x) = b^(2y - 2x) = (b^2)^(y - x)

But b^2 = ac, so that becomes:

b^(z - x) = a^(y - x) c^(y - x)

a^(y - z) b^(z - x) c^(x - y) = a^(y - z + y - x) c^(x - y + y - x)
= a^(2y - z - x) c^0
= a^(2y - z - x)

But x + z = 2y also means 2y - z - x = 0, so the exponent on a is zero too
a^(y - z) b^(z - x) c^(x - y) = a^(0) = 1

This, of course, requires a,b,c to be nonzero.

Answer 2

Hi ,

x + z = 2y-----( 1 )

x - y = y - z -----( 2 )

b² = ac -------( 3 )

LHS= a^( y-z ) ×b^(z - x ) × c^( x - y )

= a^( x - y ) × b^( z - x ) × c^( x- y )

[from ( 2 ) ]

= ( ac )^( x - y ) × b^( z - x )

= ( b² )^( x - y ) × b^ ( z - x )

[ From ( 3 ) ]

= b ^ ( 2x - 2y ) × b^( z - x )

= b^ ( 2x - 2y + z - x )

= b ^ ( x + z - 2y )

= b ^ ( 2y - 2y )

[ From ( 1 ) ]

= b^0

= 1

= RHS