Let $\sum\limits_{k = 1}^{10} {f(a + k)} = 16(2^{10} - 1),$ where the function $f$ satisfies $f(x + y) = f(x)f(y)$ for all natural numbers $x, y$ and $f(1) = 2.$ Then the natural number $a$ is

Let $f: N \rightarrow R$ be such that $f(1)=1$ and $f(1)+2 f(2)+3 f(3)+\ldots+n f(n)=n(n+1) f(n)$ for all $n \in N, n \geq 2,$ where $N$ is the set of natural numbers and $R$ is the set of real numbers. Then,the value of $f(500)$ is

Suppose $f : R \rightarrow (0, \infty)$ be a differentiable function such that $5f(x + y) = f(x) \cdot f(y), \forall x, y \in R$. If $f(3) = 320$,then $\sum_{n=0}^5 f(n)$ is equal to:

The function $f$ satisfies the functional equation $3f(x) + 2f\left( \frac{x + 59}{x - 1} \right) = 10x + 30$ for all real $x \neq 1$. The value of $f(7)$ is

Let $f$ be a function defined by $f(xy) = \frac{f(x)}{y}$ for all positive real numbers $x$ and $y$. If $f(30) = 20$,then $f(40) = $

Let $f : R - \{0, 1\} \rightarrow R$ be a function such that $f(x) + f\left(\frac{1}{1-x}\right) = 1 + x$. Then $f(2)$ is equal to: