CAT 2016 Quantitative Ability

$\mathit{A}=\left\{1,4,7,10.......20\mathrm{terms}\right\}$
$\mathit{B}=\left\{9,16.23,30......20\mathrm{terms}\right\}$
We first need to find n(A∩ B).
The ${\mathit{n}}^{\mathit{t}\mathit{h}}$ term of $\mathit{A}=1+\left(\mathit{n}-1\right)3=3\mathit{n}-2.$
The ${\mathit{m}}^{\mathit{t}\mathit{h}}$ term of $\mathit{B}=9+\left(\mathit{m}-1\right)7=7\mathit{m}+2.$
$\mathit{n}\left(\mathit{A}\cap \mathit{B}\right)$ will be equal to the number ofsolutions of $3\mathit{n}-2=7\mathit{m}+2$ , where n and m are natural numbers not more than 20.
⇒ $\mathit{n}=\frac{7\mathit{m}+4}{3}$
lf $\mathit{m}=2,\mathit{n}=6;$
$\mathit{m}=5,\mathit{n}=13;$
$\mathit{m}=8,\mathit{n}=20$
$\mathit{n}\left(\mathit{A}\bigcap \mathit{B}\right)=3$
Hence, $\mathit{n}\left(\mathit{A}\cup \mathit{B}\right)=\mathit{n}\left(\mathit{A}\right)+\mathit{n}\left(\mathit{B}\right)-\mathit{n}\left(\mathit{A}\cap \mathit{B}\right)$
$=20+20-3=37$