Let α, β and γ be three positive real numbers. Let f(x) = α x ^ 5 +...

Given $\alpha, \beta$ and $\gamma$ be three positive real numbers. Let $f\left(x\right) = \alpha{x}^{5} + \beta{x}^{3} + \gammax, x \in R$ and $g: R \to R$ be such that $g\left(f\left(x\right)\right) = x$ for all $x \in R$. If ${a}_{1}, {a}_{2}, {a}_{3}, \ldots , {a}_{n}$ be in arithmetic progression with mean zero, then the value of $f\left(g\left(\dfrac{1}{n}\mathop{∑}_{i = 1}^{n}f\left({a}_{i}\right)\right)\right)$ is equal to

Given,

Mean is zero, so $\dfrac{{a}_{1} + {a}_{2} + {a}_{3} + \ldots . . + {a}_{n}}{n} = 0$

Now this means that first and last term, second and second last and so on are equal in magnitude but opposite in sign.

Given, $f\left(x\right) = \alpha{x}^{5} + \beta{x}^{3} + \gammax$

Now finding $\mathop{∑}_{i = 1}^{n}f\left({a}_{i}\right) = \alpha\left({a}_{1}^{5} + {a}_{2}^{5} + {a}_{3}^{5} + \ldots . + {a}_{n}^{5}\right) + \beta$\left({a}_{1}^{3} + {a}_{2}^{3} + \ldots . + {a}_{n}^{3}\right) + \gamma\left({a}_{1} + {a}_{2} + \ldots . + {a}_{n}\right)$</p><p>$= 0\alpha + 0\beta + 0\gamma = 0$......equation\left(1\right)$ {as first & last, second and second last and so on are equal in magnitude and opposite in sign}

Now given $g\left(f\left(x\right)\right) = x$

$⇒ f\left(g\left(f\left(x\right)\right)\right) = f\left(x\right)$

So, $f\left(g\left(\dfrac{1}{n}\mathop{∑}_{i = 1}^{n}f\left({a}_{i}\right)\right)\right] = \dfrac{1}{n}\mathop{∑}_{i = 1}^{n}f\left({a}_{i}\right) = 0$ {from equation\left(1\right) }