Exercise 55

We have the geometric progression \(x\), \(y\), \(z\). We know that \(x\) is different from \(y\), that \(y\) is not zero, and that \(x\), \(2y\), \(3z\) form an arithmetic progression. The ratio of the geometric progression is:

  1. \( \frac{4}{3} \)

  2. \( 1 \)

  3. \( \frac{1}{3} \)

  4. \( 2 \)

  5. \( \frac{2}{3} \)

Try to solve it before checking the answer.

  1. \( \frac{1}{3} \)

Let \( r \) be the ratio of the geometric progression while \( d \) is the ratio of the aritmetic progression. Now, we have:

\( \boldsymbol{(1)} \hspace{2em} y = xr \)

\( \boldsymbol{(2)} \hspace{2em} z = xr^2 \)

\( \boldsymbol{(3)} \hspace{2em} 2y = x + d \)

\( \boldsymbol{(4)} \hspace{2em} 3z = x + 2d \)

Replacing (1) in (3) and (2) in (4):

\( \boldsymbol{(5)} \hspace{2em} 2xr = x + d \)

\( \boldsymbol{(6)} \hspace{2em} 3xr^2 = x + 2d \)

Multiplying (5) by -2:

\[ \boldsymbol{(7)} \hspace{2em} -4xr = -2x – 2d \]

Adding (6) and (7):

\[ \begin{aligned} 3xr^2 – 4xr = -x &\Rightarrow 3r^2 – 4r = -1 \hspace{2em} (\text{simplifying } x) \\[2em] &\Rightarrow 3r^2 – 4r + 1 = 0 \\[2em] &\Rightarrow r = \frac{ 4 \pm \sqrt{16 – 12} }{6} \\[2em] &\Rightarrow r = 1 \quad \text{ or } \quad r = \frac{1}{3} \end{aligned} \]
\[ \begin{aligned} &3xr^2 – 4xr = -x \\[2em] &\hspace{2em} \Rightarrow 3r^2 – 4r = -1 \\[.5em] &\hspace{4em} \text{(simplificando } x \text{)} \\[2em] &\hspace{2em}\Rightarrow 3r^2 – 4r + 1 = 0 \\[2em] &\hspace{2em}\Rightarrow r = \frac{ 4 \pm \sqrt{16 – 12} }{6} \\[2em] &\hspace{2em}\Rightarrow r = 1 \quad \text{ or } \quad r = \frac{1}{3} \end{aligned} \]

Since \( x \neq y \), by (1), we discard \( r = 1 \) and keep \( \boldsymbol{r = \frac{1}{3}} \).

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