Question
If f (x) +f (x+a) + f(x+2a) +....+ f (x+na) = constant; ∀x ϵR and a>0 and f(x) is periodic,then period of f(x), is
Answer: Option A
:
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
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:
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
Was this answer helpful ?
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