I am working through a maths text book as a hobby and am stuck on the following problem:

An arithmetic series and a geometric series have r as a common difference and the common ratio respectively. The first term of the arithmetic series is 1 and the first term of the geometric series is 2. If the fourth term of the arithmetic series is equal to the sum of the third and fourth terms of the geometric series, find the three possible values of r. When $|r| \lt 1$ find, in the form of $p + q\sqrt2$, (i) the sum to infinity of the geometric series (ii) the sum of the first ten terms of the arithmetic series.

This is my approach to the part (i), the geometric series:

The three roots are $1, -1 + \frac{1}{2}\sqrt 2$, $ -1 - \frac{1}{2}\sqrt 2$

Of these only $|-1 + \frac{1}{2}\sqrt 2| \lt 1$

$S_{\infty} = \frac{2}{1 -(-1 + \frac{\sqrt2}{2}}$

which I reduced to $\frac{1}{1 - \sqrt\frac{1}{8}}$

The book says the answer to this is $\frac{8}{7} + \frac{2}{7}\sqrt 2$

I cannot see how this is arrived at. If I apply the binomial expansion I get:

$1 + (-1)(-\frac{1}{\sqrt8}) + \frac{(-1)(-2)(-\frac{1}{\sqrt8})}{2!} +...$ so I am obviously on the wrong track.


1 Answer 1


$$\frac{1}{1-\sqrt{1/8}} = \frac{\sqrt{8}}{\sqrt{8} - 1} = \frac{\sqrt{8}( \sqrt{8} + 1)}{(\sqrt{8} - 1)(\sqrt{8} + 1)} = \frac{8 + \sqrt{8}}{8 - 1} = \frac{8}{7} + \frac{2\sqrt{2}}{7}.$$

  • $\begingroup$ How did you think of this? Is it a well-known procedure or did you just use logic? $\endgroup$
    – Steblo
    Mar 9, 2021 at 23:27
  • $\begingroup$ @Steblo This is a standard process called "rationalizing the denominator." $\endgroup$
    – heropup
    Mar 9, 2021 at 23:40

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