Convergence of Geometric series
(a) Find the sum to infinity of the GP with first term 3 and common ratio 1/2

(b) The sum to infinity of a GP is four times the first term. Find the common ratio.

(c) The sum to infinity of a GP is twice the sum of the first two terms. Find possible values of the common ratio.

Let's solve each part of this problem related to the convergence of geometric series.

### (a) Sum to Infinity of a GP

For a geometric series to converge to a finite sum, the common ratio \( r \) must satisfy \( |r| < 1 \). The sum to infinity \( S \) of a geometric series with the first term \( a \) and common ratio \( r \) is given by:

\[
S = \frac{a}{1 - r}
\]

Given:
- First term \( a = 3 \)
- Common ratio \( r = \frac{1}{2} \)

The sum to infinity is:

\[
S = \frac{3}{1 - \frac{1}{2}} = \frac{3}{\frac{1}{2}} = 3 \times 2 = 6
\]

**Answer for (a):** The sum to infinity is \( 6 \).

### (b) Common Ratio Given the Sum to Infinity

We are given that the sum to infinity of a geometric progression (GP) is four times the first term. Let's denote:
- First term as \( a \)
- Common ratio as \( r \)
- Sum to infinity as \( S \)

According to the formula:

\[
S = \frac{a}{1 - r}
\]

Given that \( S = 4a \), we can set up the equation:

\[
4a = \frac{a}{1 - r}
\]

Divide both sides by \( a \) (assuming \( a \neq 0 \)):

\[
4 = \frac{1}{1 - r}
\]

Now, solve for \( r \):

\[
1 - r = \frac{1}{4}
\]

\[
r = 1 - \frac{1}{4} = \frac{3}{4}
\]

**Answer for (b):** The common ratio is \( \frac{3}{4} \).

### (c) Common Ratio Given Sum to Infinity is Twice the Sum of First Two Terms

Let:
- First term be \( a \)
- Common ratio be \( r \)
- Sum to infinity be \( S \)

We are given:

\[
S = 2(a + ar)
\]

Substitute the sum to infinity formula:

\[
\frac{a}{1 - r} = 2(a + ar)
\]

Simplify the right side:

\[
\frac{a}{1 - r} = 2a(1 + r)
\]

Divide both sides by \( a \):

\[
\frac{1}{1 - r} = 2(1 + r)
\]

Now cross-multiply:

\[
1 = 2(1 + r)(1 - r)
\]

Simplify the right side using the difference of squares formula:

\[
1 = 2(1 - r^2)
\]

Expand and solve for \( r^2 \):

\[
1 = 2 - 2r^2
\]

\[
2r^2 = 1
\]

\[
r^2 = \frac{1}{2}
\]

Take the square root of both sides:

\[
r = \frac{1}{\sqrt{2}} \quad \text{or} \quad r = -\frac{1}{\sqrt{2}}
\]

Rationalizing the denominators, we get:

\[
r = \frac{\sqrt{2}}{2} \quad \text{or} \quad r = -\frac{\sqrt{2}}{2}
\]

**Answer for (c):** The possible values of the common ratio are \( \frac{\sqrt{2}}{2} \) and \( -\frac{\sqrt{2}}{2} \).

### Summary of Answers:
1. **(a)** The sum to infinity is \( 6 \).
2. **(b)** The common ratio is \( \frac{3}{4} \).
3. **(c)** The possible values of the common ratio are \( \frac{\sqrt{2}}{2} \) and \( -\frac{\sqrt{2}}{2} \).

Would you like a deeper explanation of any of these parts? Here are some related questions to consider:

1. What conditions are necessary for a geometric series to converge?
2. How do we derive the formula for the sum of a geometric series?
3. What happens if the common ratio \( r \) is greater than 1 or less than -1?
4. How can we use the sum of a finite geometric series in practical applications?
5. What is the relationship between the common ratio and the behavior of the geometric series?

**Tip:** Remember that for a geometric series to converge, the common ratio must lie within the interval \( -1 < r < 1 \).

Convergence of Geometric series (a) Find the sum to infinity of the GP with first term 3 and common ratio 1/2 (b) The sum to infinity of a GP is four times the first term. Find the common ratio. (c) The sum to infinity of a GP is twice the sum of the first two terms. Find possible values of the common ratio.

Solve geometric progression problems involving the sum to infinity. Learn to find the sum with the first term 3 and ratio 1/2, determine the common ratio when the sum is four times the first term, and find the ratio when the sum is twice the sum of the first two terms. Suitable for Grades 9-12.

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