If $f(x)$ is a function such that $f(x+y)=f(x)+f(y)$ and $f(1)=7$, then $\sum_{r=1}^n f(r)=$

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:

Let $R$ be the set of all real numbers and let $f$ be a function from $R$ to $R$ such that $f(x) + (x + \frac{1}{2}) f(1 - x) = 1$,for all $x \in R$. Then $2 f(0) + 3 f(1)$ is equal to

If $f: R-\{0\} \rightarrow R$ is defined by $3 f(x)+4 f\left(\frac{1}{x}\right)=\frac{2-x}{x}$, then find the value of $f(3)$.

If $g:[-2, 2] \to R$ where $g(x) = x^3 + \tan x + \left[ \frac{x^2 + 1}{P} \right]$ is an odd function,then the value of the parameter $P$ is:

Let $f$ and $g$ be functions satisfying $f(x+y)=f(x)f(y)$, $f(1)=7$ and $g(x+y)=g(xy)$, $g(1)=1$ for all $x, y \in \mathbb{N}$. If $\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)}\right) = 19607$, then $n$ is equal to: