First slide
Arithmetic progression
Question

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 

Moderate
Solution

Given that a7+a5+a8=11
a+2d+a+4d+a+7d=11
or      3a+13d=11                    (1)
Given,
a4+a2=2
a+3d+a+d=2
or           a=12d               (2)
Putting value of a from (2) in (1), we get
3(12d)+13d=117d=14d=2 and a=5
a1+a6+a7=7

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