The sum of $2n$  terms of A.P $\left\{1,5,9,13\cdots \right\}$  is greater  than sum of $n$  terms of A.P $\left\{56,58,60\cdots \right\}$  what is the smallest value $n$  can take

$S_{2n}=\frac{2n}{2}\left[2\left(1\right)+\left(2n-1\right)4\right]$

$=n\left(2+8n-4\right)=8n^2-2n$

For the second A.P $a=56,d=2$

$S_n=\frac{n}{2}\left[2\times 56+\left(n-1\right)2\right]$

$S_n=\frac{n}{2}\left[112+2n-2\right]=\frac{n}{2}\left(110+2n\right)$

Given that $8n^2-2n>\frac{n}{2}\left(110+2n\right)$

$16n^2-4n>110n+2n^2$

$14n^2>114n$

$7n^2-57n>0$

$n\left(7n-57\right)>0$

$n<0\left(or\right)n>\frac{57}{7}$

$\therefore$  The smallest  value $n$  can take is 9