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Let $f: R \rightarrow R$ be such that $f$ is injective and $f(x) f(y)=f(x+y) \quad$ for $\quad$ all $\quad x, y \in R, \quad$ if $f(x), f(y)$ and $f(z)$ are in $G P,$ then $x, y$ and $z$ are in
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The correct answer is:
AP always
Let the function, $f(x)=a^{kx}$ Which define in $f: R \rightarrow R$ and injective also.
Now, we have
$$
f(x) f(y)=f(x+y)
$$
$\Rightarrow$
$a^{k x} \cdot a^{k y}=a^{k(x+y)}$
$\Rightarrow$
$a^{k(x+y)}=a^{k(x+y)}$
$\because \quad f(x), f(y)$ and $f(z)$ are in GP
$\therefore$
$f(y)^{2}=f(x) \cdot f(z)$
$\Rightarrow$
$a^{2 k y}=a^{k x} \cdot a^{k z}$
$\Rightarrow$
$e^{2 k y}=e^{k(x+z)}$
On comparing, we get $2 k y=k(x+z) \Rightarrow 2 y=x+z$
$\Rightarrow x, y$ and $z$ are in $A P$.
Now, we have
$$
f(x) f(y)=f(x+y)
$$
$\Rightarrow$
$a^{k x} \cdot a^{k y}=a^{k(x+y)}$
$\Rightarrow$
$a^{k(x+y)}=a^{k(x+y)}$
$\because \quad f(x), f(y)$ and $f(z)$ are in GP
$\therefore$
$f(y)^{2}=f(x) \cdot f(z)$
$\Rightarrow$
$a^{2 k y}=a^{k x} \cdot a^{k z}$
$\Rightarrow$
$e^{2 k y}=e^{k(x+z)}$
On comparing, we get $2 k y=k(x+z) \Rightarrow 2 y=x+z$
$\Rightarrow x, y$ and $z$ are in $A P$.
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