In an infinite geometric series `S=(2)/(3) and t_(1)=(2)/(7)`. What is r?

Options

A. `(2)/(3)`
B. `-(4)/(7)`
C. `(2)/(7)`
D. `(4)/(7)`

Explanation

To find the common ratio \( r \) of the infinite geometric series given that the sum \( S = \frac{2}{3} \) and the first term \( t_1 = \frac{2}{7} \), we can follow these steps:

### Step 1: Write down the formula for the sum of an infinite geometric series.
The formula for the sum \( S \) of an infinite geometric series is given by:
\[ S = \frac{t_1}{1 - r} \]
where \( t_1 \) is the first term and \( r \) is the common ratio.

### Step 2: Substitute the known values into the formula.
We know \( S = \frac{2}{3} \) and \( t_1 = \frac{2}{7} \). Substituting these values into the formula gives:
\[ \frac{2}{3} = \frac{\frac{2}{7}}{1 - r} \]

### Step 3: Cross-multiply to eliminate the fraction.
Cross-multiplying yields:
\[ 2(1 - r) = 3 \cdot \frac{2}{7} \]

### Step 4: Simplify the right-hand side.
Calculating the right-hand side:
\[ 3 \cdot \frac{2}{7} = \frac{6}{7} \]
So, we have:
\[ 2(1 - r) = \frac{6}{7} \]

### Step 5: Distribute the 2 on the left-hand side.
Distributing gives:
\[ 2 - 2r = \frac{6}{7} \]

### Step 6: Isolate \( r \).
To isolate \( r \), we first move \( 2 \) to the right side:
\[ -2r = \frac{6}{7} - 2 \]
To combine the terms on the right, convert \( 2 \) into a fraction with a denominator of 7:
\[ 2 = \frac{14}{7} \]
Thus, we have:
\[ -2r = \frac{6}{7} - \frac{14}{7} = \frac{6 - 14}{7} = \frac{-8}{7} \]
Now, dividing both sides by -2 gives:
\[ r = \frac{-8/7}{-2} = \frac{8/7}{2} = \frac{8}{14} = \frac{4}{7} \]

### Final Answer:
Thus, the value of \( r \) is:
\[ \boxed{\frac{4}{7}} \]