Given $n = 20;\,\,{S_{20}} = ?$

Series $\left( 1 \right) \to 3,7,11,15,19,23,27,31,35,$

$39,43,47,$

$51,55,59…$

Series $\left( 2 \right) \to 1,6,11,16,21,26,31,36,41,$

$46,51,56,$

$61,66,71.$

the common terms between both the series are

$11,13,51,71…$

Above series froms an Arithmetic progression $(A.P)$ 

Therefore, first term$(a)=11$ and

common difference $(d)=20$

Now, ${S_n} = \frac{n}{2}\left[ {2a + \left( {n – 1} \right)d} \right]$

${S_{20}} = \frac{{20}}{2}\left[ {2 \times 11 + \left( {20 – 1} \right)20} \right]$

${S_{20}} = 10\left[ {22 + 19 \times 20} \right]$

${S_{20}} = 10 \times 402 = 4020$

$\therefore {S_{20}} = 4020$