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.