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.
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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 = xr2.
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,
for
the 1st
expression,
upon taking LCMs and simplifying:
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)]
(*)
Now,
for
the 1st
expression,
upon taking LCMs and simplifying:
we
get 1/(y
– z) – ½y
=
(y + z) / [2y(y – z)]
=
(xr + xr2)/[2xr(xr
– xr2)]
since y
= xr and z = xr2
Factoring and simplifying, we get (1
+ r)/[2xr(1 – r)]
which
is really (*) above!