An infinite geometric series has common ratio $1/8$ and sum 60.  What is the first term of the series?
Let the first term be $a$.  Because the sum of the series is 60, we have  $$60= \frac{a}{1-(1/8)} = \frac{a}{7/8} = \frac{8a}{7}.$$Therefore, $a=\frac{7}{8}\cdot60=\boxed{\frac{105}{2}}$.
\frac{105}{2}
Level 4
Algebra
Problem: An infinite geometric series has common ratio $1/8$ and sum 60.  What is the first term of the series?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the first term 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, a is the first term, and r is the common ratio. 

In this case, we know the sum S is 60 and the common ratio r is 1/8. We can substitute these values into the formula:

60 = a / (1 - 1/8).

Next, we simplify the denominator:

1 - 1/8 = 7/8.

Now, we can rewrite the equation:

60 = a / (7/8).

To eliminate the fraction, we multiply both sides by 7/8:

60 * (7/8) = a.

Calculating the left side gives:

60 * 7 = 420, and then dividing by 8 results in:

420 / 8 = 52.5.

Thus, we find that a = 52.5. 

To express this as a fraction, we convert 52.5 into a fraction:

52.5 = 52.5 * (2/2) = 105/2.

Therefore, the first term of the series is $\frac{105}{2}$.<|END_OF_LATENT|> The final answer is $\frac{105}{2}$.