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 ½y1/(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!