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(A) Given the functional equation $f(x)f(y) = f(x+y)$.Since $f$ is injective, the solution to this Cauchy functional equation is of the form $f(x) = a

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$AP$ always

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(A) Given the functional equation $f(x)f(y) = f(x+y)$.Since $f$ is injective, the solution to this Cauchy functional equation is of the form $f(x) = a^{kx}$ for some constant $a > 0, a 
eq 1$ and $k 
eq 0$.Given that $f(x), f(y),$ and $f(z)$ are in $GP$, we have $(f(y))^2 = f(x)f(z)$.Substituting $f(x) = a^{kx}$, we get $(a^{ky})^2 = a^{kx} \cdot a^{kz}$.This simplifies to $a^{2ky} = a^{k(x+z)}$.Since $a 
eq 1$ and $k 
eq 0$, we equate the exponents: $2ky = k(x+z)$.Dividing by $k$, we get $2y = x+z$.This condition implies that $x, y,$ and $z$ are in $AP$.

Let $f: R \rightarrow R$ be such that $f$ is injective and $f(x)f(y) = f(x+y)$ for all $x, y \in R$. If $f(x), f(y),$ and $f(z)$ are in $GP$, then $x, y,$ and $z$ are in:

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