Let's solve the problem step by step:

(a) Find the common ratio.

In a geometric series, the common ratio $r$ is the factor by which we multiply each term to get the next term. Given the series:

$512 - 256 + 128 - 64 + \dots$

We can find the common ratio by dividing the second term by the first term:

$r = \frac{-256}{512} = -\frac{1}{2}$

So, the common ratio is:

$r = -\frac{1}{2}$

(b) Find the twelfth term of the series.

The general term of a geometric series can be written as:

$T_n = a \cdot r^{n-1}$

Where:

• $a = 512$ (the first term),
• $r = -\frac{1}{2}$,
• $n = 12$.

So, the twelfth term is:

$T_{12} = 512 \cdot \left(-\frac{1}{2}\right)^{12-1} = 512 \cdot \left(-\frac{1}{2}\right)^{11}$

Calculating:

$T_{12} = 512 \cdot \left(-\frac{1}{2}\right)^{11} = 512 \cdot -\frac{1}{2048}$

$T_{12} = \frac{-512}{2048} = -\frac{1}{4}$

So, the twelfth term is:

$T_{12} = -\frac{1}{4}$

(c) Find the exact value of the sum of the infinite series.

The sum $S$ of an infinite geometric series is given by the formula:

$S = \frac{a}{1 - r}$

Where:

• $a = 512$,
• $r = -\frac{1}{2}$.

Substitute the values:

$S = \frac{512}{1 - \left(-\frac{1}{2}\right)} = \frac{512}{1 + \frac{1}{2}} = \frac{512}{\frac{3}{2}} = 512 \cdot \frac{2}{3} = \frac{1024}{3}$

So, the exact value of the sum of the infinite series is:

$S = \frac{1024}{3}$

Final Answers:

1. (a) The common ratio is $r = -\frac{1}{2}$.
2. (b) The twelfth term is $T_{12} = -\frac{1}{4}$.
3. (c) The exact value of the sum of the infinite series is $S = \frac{1024}{3}$.

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Would you like more details or have any questions on any step?

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