Same current (I) flows through different resistances, when these are joined in series, as shown in the figure.Let $\mathrm{R}$ be the combined resistance, then
$\mathrm{V}=\mathrm{IR}$
Also, ${\mathit{V}}_1=\mathit{I}{\mathit{R}}_1,{\mathit{V}}_2=\mathit{I}{\mathit{R}}_2,{\mathit{V}}_3=\mathit{I}{\mathit{R}}_3$
$\because \mathrm{V}={\mathrm{V}}_1+{\mathrm{V}}_2+{\mathrm{V}}_3$
$\therefore \mathrm{IR}={\mathrm{IR}}_1+{\mathrm{IR}}_2+{\mathrm{IR}}_3$
$\Rightarrow \mathrm{IR}=\mathrm{I}\left({\mathrm{R}}_1+{\mathrm{R}}_2+{\mathrm{R}}_3\right)$
$\therefore \mathrm{R}={\mathrm{R}}_1+{\mathrm{R}}_2+{\mathrm{R}}_3$
Now, ${\mathit{R}}_1=6\mathit{\Omega },{\mathit{R}}_2=9\mathit{\Omega },{\mathit{R}}_3=18\mathit{\Omega }$
In parallel combination,
$\frac{1}{\mathrm{R}}=\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}+\frac{1}{{\mathrm{R}}_3}$
$\Rightarrow \frac{1}{\mathrm{R}}=\frac{1}{6}+\frac{1}{9}+\frac{1}{18}=\frac{3+2+1}{18}$
$=\frac{6}{18}=\frac{1}{3}$
$\Rightarrow \frac{1}{\mathrm{R}}=\frac{1}{3}$
$\Rightarrow \mathrm{R}=3\mathit{\Omega }$