Let the first term be $a$ and the common ratio be $r$.
We are given that the third term is 12, so we have $ar^2 = 12$.
We are also given that the fourth term is 16, so we have $ar^3 = 16$.
Dividing the second equation by the first equation, we get $\frac{ar^3}{ar^2} = \frac{16}{12}$.
Canceling out the $a$ and simplifying, we have $r = \frac{4}{3}$.
Substituting this value into the first equation, we have $a(\frac{4}{3})^2 = 12$.
Simplifying, we get $\frac{16a}{9} = 12$.
Dividing both sides by $\frac{16}{9}$, we get $a = \frac{27}{4}$.
Therefore, the value of the first term in the sequence is $\boxed{\frac{27}{4}}$.The answer is: \frac{27}{4}
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
In a geometric sequence, the third term is 12 and the fourth term is 16. What is the value of the first term in the sequence?

### Answer:
