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.

This is how I have calculated:

Let $u_{n}$ refer to the nth member of a series. Then for the arithmetic series the 4th term is $1 + 3r$.

If $S_{n}$ refers to the sum to n of a series then $S_{4} - S_{2}$ will give $u_{3} + u_{4}$ of the geometric series.

$S_{4} = \frac{2(1 - r^4)}{1 -r}$ and $S_{2} = \frac{2(1 - r^2)}{1 -r}$

$S_{4} - S_{2} = \frac{2r^2 - 2r^4}{1-r}$

$(1 + 3r) = \frac{2r^2 - 2r^4}{1-r}$

$(1 + 3r)(1-r) = 2r^2 - 2r^4$

$\implies 2r^4 - 5r^2 + 2r + 1 = 0$

Now my textbook hasn't got as far as solving polynomials higher than 3 yet - and I cannot do it - and I'm wondering if there is another way of approaching this problem.

  • $\begingroup$ Clearly, $r=1$ is a solution $\endgroup$ Mar 9 at 11:37
  • $\begingroup$ Yes, for solving biquadratic it is a very useful technique to just put common values such as $$x=0,\pm1,\pm2,\pm3$$ This way a factor can be known and then one can solve a cubic after that. $\endgroup$
    – GuyEternal
    Aug 3 at 17:39

The terms of GP are $2,2r,2r^2,2r^3,\ldots$ So we can directly write

$$1+3r=2r^2+2r^3$$ $$\Rightarrow 2r^3+2r^2-3r-1=0$$

which indicates that there is mistake made in simplification in OP. EDIT : There is no mistake in simplification as pointed by @DavidK below in comments since $$2r^4-5r^2+2r+1=(r-1)(2r^3+2r^2-3r-1)$$


Clearly $1$ is a root, so the above factorizes as $$(r-1)(2r^2+4r+1)=0$$

The second factor is a quadratic whose roots can be found using quadratic formula.

  • $\begingroup$ Original post or poster, would be clear from context :) $\endgroup$
    – cosmo5
    Mar 9 at 12:01
  • 1
    $\begingroup$ OP did not make any mistake in the algebraic manipulations. $(2 r^3 + 2 r^2 - 3 r - 1) (r - 1) = 2r^4 - 5r^2 + 2r + 1.$ Since $r=1$ is a root of $2 r^3 + 2 r^2 - 3 r - 1$, this does not even introduce an extraneous solution. The "mistake" was in not recognizing the factor of $r-1$ in $2r^4-2r^2$ -- or even earlier, using the excessively complex partial-sum formulas rather than just adding the two terms $2r^2$ and $2r^3$ directly. $\endgroup$
    – David K
    Mar 9 at 12:51
  • $\begingroup$ @Steblo, $2r^2+4r+1=0$ has roots $-2 \pm \sqrt{2}$. Note that $(2r-2)(r-1)=2r^2-4r+\color{red}{2}$. $\endgroup$
    – cosmo5
    Mar 9 at 13:01

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