We are given a geometric sequence $\{a_n\}$ where $a_1 = 3$. We also know that $4a_1, 2a_2, a_3$ forms an arithmetic sequence. We are asked to find the value of $a_3 + a_4 + a_5$.

1. Problem Description

We are given a geometric sequence $\{a_n\}$ where $a_1 = 3$. We also know that $4a_1, 2a_2, a_3$ forms an arithmetic sequence. We are asked to find the value of $a_3 + a_4 + a_5$.

2. Solution Steps

Since $\{a_n\}$ is a geometric sequence, we have $a_n = a_1 \cdot r^{n-1}$, where $r$ is the common ratio. Therefore, $a_2 = a_1 \cdot r = 3r$ and $a_3 = a_1 \cdot r^2 = 3r^2$.

Since $4a_1, 2a_2, a_3$ is an arithmetic sequence, we have

$$2(2a_2) = 4a_1 + a_3$$

$$4a_2 = 4a_1 + a_3$$

Substituting $a_1 = 3$, $a_2 = 3r$, and $a_3 = 3r^2$ into the equation, we get

$$4(3r) = 4(3) + 3r^2$$

$$12r = 12 + 3r^2$$

$$3r^2 - 12r + 12 = 0$$

$$r^2 - 4r + 4 = 0$$

$$(r-2)^2 = 0$$

$$r = 2$$

Thus, $a_1 = 3, a_2 = 3(2) = 6, a_3 = 3(2^2) = 12, a_4 = 3(2^3) = 24, a_5 = 3(2^4) = 48$.

Therefore, $a_3 + a_4 + a_5 = 12 + 24 + 48 = 84$.

3. Final Answer

84