If f : ℝ → ℝ is such that f ( x + y ) = f ( x ) + f ( y ) for all x , y ∈ ℝ ,…

We have $f:\mathrm{ℝ}\to \mathrm{ℝ}$ is such that

$f(x+y)=f\left(x\right)+f\left(y\right)$ for all $x,y\in \mathrm{ℝ}, f\left(1\right)=7$.

Now,

$f\left(2\right)=f\left(1+1\right)=f\left(1\right)+f\left(1\right)=2·7$

Similarly,

$f\left(3\right)=f\left(2+1\right)=f\left(2\right)+f\left(1\right)=3·7$

$f\left(4\right)=f\left(3+1\right)=f\left(3\right)+f\left(1\right)=4·7$

$$\begin{gathered} ⋮ ⋮ ⋮ \\ ⋮ ⋮ ⋮ \\ ⋮ ⋮ ⋮ \end{gathered}$$

$f\left(n-1\right)=7·\left(n-1\right)$

$f\left(n\right)=n·7$

Therefore,

 $\sum _{r=1}^{n}f\left(r\right)=14112$

$\Rightarrow f\left(1\right)+f\left(2\right)+f\left(3\right)+...+f\left(n\right)=14112$

$\Rightarrow 1·7+2·7+3·7+....+\left(n-1\right)·7+n·7=14112$

$\Rightarrow 7\left(1+2+3+....+\left(n-1\right)+n\right)=14112$

$\Rightarrow 7\left\{\frac{n\left(n+1\right)}{2}\right\}=14112$

$\Rightarrow n\left(n+1\right)=\frac{14112\times 2}{7}$

$\Rightarrow n\left(n+1\right)=4032$

$\Rightarrow n\left(n+1\right)=63\left(63+1\right)$

$\Rightarrow n=63$.