If f (x) +f (x+a) + f(x+2a) +....+ f (x+na) = constant; ∀xϵR and a>0 and f(x) is periodic,then period of f(x), is
A
(n+1) a
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B
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C
na
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D
e^{na}
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Solution
The correct option is A (n+1) a f(x)+f(x+a)+f(x+2a)+....+f(x+na)=k replacing x=x+a f(x+a)+f(x+2a)+....+f(x+(n+1)a)=k On subtracting second equation from first one - f(x)−f(x+(n+1)a)=0f(x)=f(x+(n+1)a∴T=(n+1)a