An infinite geometric series has first term $328$ and a sum of $2009$. What is its common ratio?
Since this is an infinite geometric series, we have $\frac{328}{1-r} = 2009$. Solving for $r$, we find that $r = \boxed{\frac{41}{49}}$.
\frac{41}{49}
Level 5
Algebra
Problem: An infinite geometric series has first term $328$ and a sum of $2009$. What is its common ratio?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the common ratio of the infinite geometric series, we start with the formula for the sum of an infinite geometric series, which is given by:

S = a / (1 - r)

where S is the sum of the series, a is the first term, and r is the common ratio. In this case, we know that the first term a is 328 and the sum S is 2009. 

Substituting the known values into the formula, we have:

2009 = 328 / (1 - r)

Next, we need to isolate the common ratio r. We can start by multiplying both sides of the equation by (1 - r):

2009(1 - r) = 328

Expanding this gives:

2009 - 2009r = 328

Now, we can isolate the term involving r by moving 2009 to the other side:

-2009r = 328 - 2009

Calculating the right side:

328 - 2009 = -1681

Thus, we have:

-2009r = -1681

Dividing both sides by -2009 gives:

r = -1681 / -2009

This simplifies to:

r = 1681 / 2009

Next, we can simplify the fraction. We find the greatest common divisor (GCD) of 1681 and 2009. The prime factorization of 1681 is 41^2, and the prime factorization of 2009 is 7^2 * 41. The GCD is 41.

Now, we divide both the numerator and the denominator by their GCD:

r = (1681 / 41) / (2009 / 41)

This simplifies to:

r = 41 / 49

Thus, the common ratio of the infinite geometric series is confirmed to be 41/49.<|END_OF_LATENT|> The final answer is $\frac{41}{49}$.