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

Given that a3+a5+a8=11⇒a+2d+a+4d+a+7d=11⇒3a+13d=11 Given a4+a2=−2⇒a+3d+a+d=−2⇒a=−1−2dPutting value of ‘a’ from (2) in (1), we get3(−1−2d)+13d=11⇒7d=14⇒d=2 and a=−5⇒a1+a6+a7=7