$\therefore$ We know that sum of equidistant term from end & beginning is same in any $\mathrm{A}.\mathrm{P}.$$\Rightarrow a_1+a_{16}{=a}_2+a_{15}=a_3+a_{14}{=a}_4+a_{13}=a_5+a_{12}=\dots =a_7+a_{10}$

$\because a_1+a_4{+a}_7+a_{10}+a_{13}+a_{16}=114$

$\Rightarrow 3\left(a_1+a_{16}\right)=114\Rightarrow a_1+a_{16}=38$

Then,

$a_1+a_6{+a}_{11}+a_{16}$

$=2\left(a_1+a_{16}\right)=76$