If $f(x) = \frac{x}{x - 1} = \frac{1}{y}$,then $f(y) = $

Find $\sum_{t=1}^{39} f(t)$ if $f: R \rightarrow R$ is defined as $f(x+y)=f(x)+f(y)$ for all $x, y \in R$ and $f(1)=7$.

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

If $\phi (x) = a^x$,then $\{ \phi (p) \} ^3$ is equal to

Let $f: R \rightarrow R$ be defined as $f(x+y)+f(x-y)=2 f(x) f(y)$ and $f\left(\frac{1}{2}\right)=-1$. Then,the value of $\sum_{k=1}^{20} \frac{1}{\sin (k) \sin (k+f(k))}$ is equal to:

$f: R \rightarrow R$ is defined by $f(x+y)=f(x)+12y, \forall x, y \in R$. If $f(1)=6$,then $\sum_{r=1}^n f(r)=$