Let α , β and γ be three positive real numbers. Let f x = α x 5 + β x 3 + γ x , x ∈ R and g : R → R be such that g f x = x for all x ∈ R . If a 1 , a 2 , a 3 , … , a n be in arithmetic progression with mean zero, then the value of f g 1 n ∑ i = 1 n f a i is equal to

Step-by-step solution

Given α , β and γ be three positive real numbers. Let f x = α x 5 + β x 3 + γ x , x ∈ R and g : R → R be such that g f x = x for all x ∈ R . If a 1 , a 2 , a 3 , … , a n be in arithmetic progression with mean zero, then the value of f g 1 n ∑ i = 1 n f a i is equal to Given, Mean is zero, so a 1 + a 2 + a 3 + … . . + 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,