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If a, b, c are three consecutive terms of an A.P. and x, y, z are three consecutive terms of a G.P. then prove that xb−c × yc−a × za−b = 1 .

If a, b, c are three consecutive terms of an A.P. and x, y, z are three consecutive terms of a G.P. then prove that xb−c × yc−a × za−b = 1 .
Solution :
Since a, b and c are in A.P,
b - a  = c - b  = d (common difference)
We need to prove,
xb−c × yc−a × za−b = 1 
Let us try to convert the powers in terms of one variable.
2b  = c + a - a + a
2b  = c - a + 2a
2(b - a)  = c - a
2d  = c - a 
If c - b  = d, then b - c = -d
If b - a  = d, then a - b = -d
L.H.S
xb−c × yc−a × za−b  =  x−d × y2d × z−d   ---(1)
y = √xz
By applying the value of y in (1)
 =  x−d × (√xz)2d × z−d  
 =  x−d × (xz)d × z−d  
  =  x−d + d  z-d + d
  =  1

Hence proved.

If a, b, c are in A.P. then show that 3a, 3b, 3c are in G.P.

If a, b, c are in A.P. then show that 3a, 3b, 3c are in G.P.
Solution :
If a, b, c are in A.P, then b = (a + c)/2
In G.P,  b = √ac
To prove that 3a, 3b, 3c are in G.P.
3b  = √(3a ⋅ 3c) 
3b  = (3a + c)1/2
3b  = 3(a+c)/2
b = (a + c)/2
Hence 3a, 3b, 3c are in G.P.