**9/03/2014****
**If
x, y and z are 3 consecutive terms of a geometric sequence, prove
that 1/(y – x), ½y
and
1/(y – z) are terms in an arithmetic sequence.

**Scroll
down for the solution.**

**9/03/2014
**Solution.
**Since
x****,
y and z are 3 consecutive terms of a geometric sequence, ****let
y = xr where r is the common ratio, so that z = xr**^{2}**.
To show that ****1/(y
– x), ½y****
****and
1/(y – z) are terms in an arithmetic sequence, we shall
demonstrate that ****the
common difference between ½y****
****and
1/(y – x)****
****is
the same as ****the
common difference between 1/(y – z) and ½y****.Now,
**

**we
get****
½y****
– ****1/(y
– x)****
**

**=
(-x – y)/[2y( y – x)] = (-x – xr) / [2xr(xr –
x)] using ****y
= xr from above.**

**=
****-(1
+ r)/[2xr(r – 1)]****
****(*)**

**we
get ****1/(y
– z) – ½y **

=
(y + z) / [2y(y – z)]

**=
(xr + xr**^{2}**)/[2xr(xr
– xr**^{2}**)]
since ****y
= xr and z = xr**^{2}**
Factoring and simplifying, we get **