A.P

Let the first term and the common difference be :

$a_{1},\ d_{1}\ and\ a_{2}\ and\ d_{2}\ respectively$ $\frac {2a_{1}+(n-1)d_{1} )} {a_{2} + (n-1) d_{2} } = \frac {7n +1} {4n + 27 }$

$We\ are\ supposed\ to\ find\ the \frac { a_{1} + 10d_{2} } { a_{2} + 10 d_{2} }$

$therefore\ substitute\ n = 21 \ we\ will\ get : \quad$

$\rightarrow \frac {2a_{1} + 20d_{1} } { 2a_{2} + 20d_{2} }$ $The\ Answer\ drops\ simply\ to:$ $\frac {148}{111} = \frac {4} {3} = \huge \boxed { \color{#3D99F6}{ 4*3=12} }$