Let $$\alpha, \beta$$ and $$\gamma$$ be three positive real numbers. Let $$f(x) = \alpha x^5 + \beta x^3 + \gamma x$$, $$x \in \mathbb{R}$$ and $$g: \mathbb{R} \to \mathbb{R}$$ be such that $$g(f(x)) = x$$ for all $$x \in \mathbb{R}$$. If $$a_1, a_2, a_3, \ldots, a_n$$ be in arithmetic progression with mean zero, then the value of $$f\left(g\left(\frac{1}{n}\sum_{i=1}^{n} f(a_i)\right)\right)$$ is equal to

We need to find $$f\left(g\left(\frac{1}{n}\sum_{i=1}^{n} f(a_i)\right)\right)$$ where $$a_1, a_2, \ldots, a_n$$ are in AP with mean zero.

Step 1: Understand the given functions.

$$f(x) = \alpha x^5 + \beta x^3 + \gamma x$$ where $$\alpha, \beta, \gamma > 0$$.

$$g(f(x)) = x$$ for all $$x \in \mathbb{R}$$, so $$g = f^{-1}$$ ($$g$$ is the inverse function of $$f$$).

Step 2: Observe that $$f$$ is an odd function.

$$f(-x) = \alpha(-x)^5 + \beta(-x)^3 + \gamma(-x) = -\alpha x^5 - \beta x^3 - \gamma x = -f(x)$$

So $$f$$ is an odd function: $$f(-x) = -f(x)$$.

Step 3: Use the AP with mean zero property.

Since $$a_1, a_2, \ldots, a_n$$ are in AP with mean zero:

$$\frac{1}{n}\sum_{i=1}^{n} a_i = 0$$

The terms are symmetric about zero, meaning they can be paired as $$a_i$$ and $$a_{n+1-i} = -a_i$$.

Step 4: Evaluate $$\sum f(a_i)$$.

Since the $$a_i$$ are symmetric about 0 and $$f$$ is odd:

$$f(a_i) + f(a_{n+1-i}) = f(a_i) + f(-a_i) = f(a_i) - f(a_i) = 0$$

Each pair sums to zero. If $$n$$ is odd, the middle term is $$a_{(n+1)/2} = 0$$, and $$f(0) = 0$$.

Therefore: $$\sum_{i=1}^{n} f(a_i) = 0$$

Step 5: Compute the final expression.

$$\frac{1}{n}\sum_{i=1}^{n} f(a_i) = \frac{0}{n} = 0$$

$$g(0) = ?$$ Since $$f(0) = 0$$, and $$g = f^{-1}$$, we get $$g(0) = 0$$.

$$f(g(0)) = f(0) = 0$$

The correct answer is Option A: $$0$$