Consider an A.P.${\mathrm{a}}_1,{\mathrm{a}}_2,{\mathrm{a}}_3,\dots$ such that ${\mathrm{a}}_3+{\mathrm{a}}_5+{\mathrm{a}}_8=11$ and ${\mathrm{a}}_4+{\mathrm{a}}_2=-2$ then the value of ${\mathrm{a}}_1+{\mathrm{a}}_6+{\mathrm{a}}_7$ is 

Given that ${\mathrm{a}}_7+{\mathrm{a}}_5+{\mathrm{a}}_8=11$
$\Rightarrow \mathrm{a}+2\mathrm{d}+\mathrm{a}+4\mathrm{d}+\mathrm{a}+7\mathrm{d}=11$
or      $3\mathrm{a}+13\mathrm{d}=11 \left(1\right)$
Given,
${\mathrm{a}}_4+{\mathrm{a}}_2=-2$
$\Rightarrow \mathrm{a}+3\mathrm{d}+\mathrm{a}+\mathrm{d}=-2$
or           $\mathrm{a}=-1-2\mathrm{d} \left(2\right)$
Putting value of a from (2) in (1), we get
$3(-1-2\mathrm{d})+13\mathrm{d}=11\Rightarrow 7\mathrm{d}=14\Rightarrow \mathrm{d}=2\text{ and }\mathrm{a}=-5$
$\Rightarrow {\mathrm{a}}_1+{\mathrm{a}}_6+{\mathrm{a}}_7=7$