Consider the infinite geometric series 512-256 + 128 - 64 +•..
(a) Find the common ratio, giving your answer as a fraction in its simplest form.
(b) Find the twelfth term of this series.
(c) Find the exact value of the sum of the infinite series.

We are given the infinite geometric series:  
\[ 512 - 256 + 128 - 64 + \cdots \]

Let's address each part of the question step by step:

### (a) Find the common ratio
In a geometric series, the common ratio \( r \) is the factor by which each term is multiplied to get the next term.

To find the common ratio:
\[
r = \frac{\text{second term}}{\text{first term}} = \frac{-256}{512} = -\frac{1}{2}
\]

So, the common ratio is:
\[
r = -\frac{1}{2}
\]

### (b) Find the twelfth term of this series
The general term of a geometric series is given by:
\[
T_n = a \cdot r^{n-1}
\]
where:
- \( a = 512 \) (the first term),
- \( r = -\frac{1}{2} \),
- \( n = 12 \) (we need the 12th term).

Substituting into the formula:
\[
T_{12} = 512 \cdot \left( -\frac{1}{2} \right)^{12-1} = 512 \cdot \left( -\frac{1}{2} \right)^{11}
\]
Since \( (-1)^{11} = -1 \), this becomes:
\[
T_{12} = 512 \cdot -\frac{1}{2^{11}} = 512 \cdot -\frac{1}{2048}
\]
Simplifying:
\[
T_{12} = -\frac{512}{2048} = -\frac{1}{4}
\]

So, the twelfth term is:
\[
T_{12} = -\frac{1}{4}
\]

### (c) Find the exact value of the sum of the infinite series
The sum of an infinite geometric series is given by the formula:
\[
S_{\infty} = \frac{a}{1 - r}
\]
where \( |r| < 1 \).

Substituting the values \( a = 512 \) and \( r = -\frac{1}{2} \):
\[
S_{\infty} = \frac{512}{1 - \left( -\frac{1}{2} \right)} = \frac{512}{1 + \frac{1}{2}} = \frac{512}{\frac{3}{2}} = 512 \cdot \frac{2}{3}
\]
Simplifying:
\[
S_{\infty} = \frac{1024}{3}
\]

Thus, the exact value of the sum of the infinite series is:
\[
S_{\infty} = \frac{1024}{3}
\]

### Final Answers:
(a) The common ratio is \( -\frac{1}{2} \).  
(b) The twelfth term is \( -\frac{1}{4} \).  
(c) The sum of the infinite series is \( \frac{1024}{3} \).

---

Do you have any questions or would you like more details on any part? Here are 5 related questions you can explore:

1. How does the common ratio affect the convergence of an infinite geometric series?
2. How would you find the sum of the first 12 terms of this series?
3. What happens to the sum of the series if the common ratio is greater than 1?
4. How can the formula for the sum of an infinite geometric series be derived?
5. What is the difference between a finite and infinite geometric series in terms of summation?

**Tip**: Always check that \( |r| < 1 \) before applying the sum formula for infinite geometric series, as this condition ensures convergence!

Consider the infinite geometric series 512 - 256 + 128 - 64 + •.. (a) Find the common ratio, giving your answer as a fraction in its simplest form. (b) Find the twelfth term of this series. (c) Find the exact value of the sum of the infinite series.

Learn how to find the common ratio, twelfth term, and sum of an infinite geometric series. This example explores a series with the first term 512 and a common ratio of -1/2. Suitable for students in Grades 10-12.

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