The first three terms of a geometric series are $x^2$, $5x - 8$, and $x + 8$, where $x \in \mathbb{R}$. Use the common ratio to show that $x^3 - 17x^2 + 80x - 64 = 0$. If $f(x) = x^3 - 17x^2 + 80x - 64$, $x \in \mathbb{R}$, show that $f(1) = 0$, and find another value of $x$ for which $f(x) = 0$. In the case of one of the values of $x$ from part (b), the terms in part (a) will generate a geometric series with a finite sum to infinity. Find this value of $x$ and hence find the sum to infinity.

Leaving Cert Mathematics Sequences & Series: The first three terms of a geome

The first three terms of a geometric series are $x^2$, $5x - 8$, and $x + 8$, where $x \in \mathbb{R}$ - Leaving Cert Mathematics - Question 2 - 2018

Question 2

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The first three terms of a geometric series are $x^2$, $5x - 8$, and $x + 8$, where $x \in \mathbb{R}$. Use the common ratio to show that $x^3 - 17x^2 + 80x - 64 = 0... show full transcript

Worked Solution & Example Answer:The first three terms of a geometric series are $x^2$, $5x - 8$, and $x + 8$, where $x \in \mathbb{R}$ - Leaving Cert Mathematics - Question 2 - 2018

Step 1

Show that $x^3 - 17x^2 + 80x - 64 = 0$

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Answer

To find the common ratio, we set up the ratio between consecutive terms:

$r = \frac{5x - 8}{x^2} = \frac{x + 8}{5x - 8}$

Cross-multiplying gives:

$(5x - 8)^2 = x^2 (x + 8)$

Expanding both sides:

$(25x^2 - 80x + 64) = (x^3 + 8x^2)$

Rearranging the equation leads to:

$x^3 - 17x^2 + 80x - 64 = 0$

Step 2

Show that $f(1) = 0$

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Answer

We evaluate:

$f(1) = 1^3 - 17(1)^2 + 80(1) - 64$

Calculating:

$f(1) = 1 - 17 + 80 - 64 = 0$

So, $f(1) = 0$ shows that $x = 1$ is a root.

To find another value of $x$ for which $f(x) = 0$ , we can factor $f(x)$ as $f(x) = (x - 1)(x^2 - 16)$ . Solving $x^2 - 16 = 0$ gives:

$x^2 = 16 \Rightarrow x = 4 \text{ or } x = -4.$

Step 3

Find this value of $x$ and hence find the sum to infinity

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Answer

From part (b), we have $x = 8$ as suitable since one root generates a valid geometric series.

For the series, $S_\infty = \frac{a}{1 - r},$ where $a$ is the first term and $r$ is the common ratio. With $x = 8$ :

First term: $a = 8^2 = 64$ .

Common ratio: $r = \frac{5(8) - 8}{8^2} = \frac{32}{64} = \frac{1}{2}.$

Thus: $S_\infty = \frac{64}{1 - \frac{1}{2}} = \frac{64}{\frac{1}{2}} = 128.$

The first three terms of a geometric series are $x^2$, $5x - 8$, and $x + 8$, where $x \in \mathbb{R}$ - Leaving Cert Mathematics - Question 2 - 2018

Worked Solution & Example Answer:The first three terms of a geometric series are $x^2$, $5x - 8$, and $x + 8$, where $x \in \mathbb{R}$ - Leaving Cert Mathematics - Question 2 - 2018

Show that $x^3 - 17x^2 + 80x - 64 = 0$

Show that $f(1) = 0$

Find this value of $x$ and hence find the sum to infinity

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