$$
\text { Given } f: R \rightarrow R \text { is defined as }
$$
$$
f(x+y)=f(x)+f(y), \forall x, y \in R \text { and } f(1)=5
$$

To Find $\sum_{r=1}^n f(r)=$ ?
Since, $f(x+y)=f(x)+f(y)$
Let $x=y=1$
$$
\begin{aligned}
f(1+1) & =f(\mathrm{l})+f(\mathrm{l})=5+5 \\
f(2) & =10
\end{aligned}
$$

Again in Eq. (i), put $x=2, y=1$
$$
\begin{aligned}
f(2+1) & =f(2)+f(1) \\
f(3) & =10+5 \\
f(3) & =15
\end{aligned}
$$

Similarly, we can find $f(4)=15+5=20$
$$
\begin{aligned}
f(5) & =25, \ldots f(n)=5 n \\
\sum_{r=1}^n f(r) & =f(1)+f(2)+\ldots+f(n) \\
& =5+10+15+\ldots+5 n
\end{aligned}
$$
which forms an AP of $n$-terms with first term (a) $=5$ and common difference $(d)=5$
$$
\begin{aligned}
& =\frac{n}{2}[2(5)+(n-1) 5] \\
& =\frac{n}{2}[10+5 n-5]=\frac{n}{2}[5 n+5] \\
\sum_{r=1}^n f(r) & =\frac{5 n(n+1)}{2}
\end{aligned}
$$