The third and fourth terms of a geometric sequence are 12 and 16, respectively. What is the first term of the sequence?
The common ratio of the geometric sequence is $\frac{16}{12} = \frac{4}{3}$. Thus, the $n^{\text{th}}$ term of the sequence is given by $a\left(\frac{4}{3}\right)^{n-1}$, where $a$ is the first term of the sequence. Since the third term is 12, we plug in $n=3$ to get $a\left(\frac{4}{3}\right)^2 = 12$. Solving, we have $a\left(\frac{16}{9}\right) = 12 \Rightarrow a = 12\left(\frac{9}{16}\right) = \boxed{\frac{27}{4}}$.
\frac{27}{4}
Level 4
Algebra
Problem: The third and fourth terms of a geometric sequence are 12 and 16, respectively. What is the first term of the sequence?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the first term of the geometric sequence, denote the first term as a and the common ratio as r. The third term can be expressed as ar^2 and the fourth term as ar^3. Given that ar^2 = 12 and ar^3 = 16, we can set up the following equations:

1. ar^2 = 12
2. ar^3 = 16

To eliminate a, divide the second equation by the first:

(ar^3) / (ar^2) = 16 / 12

This simplifies to:

r = 16 / 12 = 4 / 3

Now that we have the common ratio r, substitute r back into the first equation to find a:

ar^2 = 12
a(4/3)^2 = 12
a(16/9) = 12

To isolate a, multiply both sides by the reciprocal of 16/9:

a = 12 * (9/16)
a = 108 / 16
a = 27 / 4

Thus, the first term of the sequence is 27/4.<|END_OF_LATENT|> The final answer is $\frac{27}{4}$.