If $\overrightarrow{A}+\overrightarrow{B}=\overrightarrow{R}\text{and}2\overrightarrow{A}+\overrightarrow{B}$ is perpendicular to $\overrightarrow{B}$  then     

Given $\left(2\text{A}+\text{B}\right)\text{is}\perp \text{to}\text{B}$

$\left(2\text{A}+\text{B}\right).\text{B}=0$

$2\text{AB}+{\text{B}}^2=0\to \left(1\right)$

Given $\text{R}=\text{A}+\text{B}$

$\text{B}=\text{R}-\text{A}$

$2\text{RA}-2{\text{A}}^{\text{2}}+{\text{R}}^{\text{2}}+{\text{A}}^{\text{2}}-2\text{RA}=0$

$={\text{R}}^{\text{2}}-{\text{A}}^{\text{2}}$

${\text{R}}^{\text{2}}={\text{A}}^{\text{2}}$

$\text{R}=\text{A}$