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Arithmetic progression

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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 

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Solution

a+2d+a+4d+a+7d=113a+13d=11 (1) 

a+3d+a+d=2a+2d=12 From (1) and (2) d=2, a=-5

a1+a6+a7=3a+11d=7


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