Consider an A.P.a1,a2,a3,… such that a3+a5+a8=11 and a4+a2=−2 then the value of a1+a6+a7 is
-8
5
7
9
Given that a7+a5+a8=11⇒a+2d+a+4d+a+7d=11or 3a+13d=11 (1)Given,a4+a2=−2⇒a+3d+a+d=−2or a=−1−2d (2)Putting value of a from (2) in (1), we get3(−1−2d)+13d=11⇒7d=14⇒d=2 and a=−5⇒a1+a6+a7=7