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.