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Let a1,a2,a3, be an A.P. such that a3+a5+a8=11 and a4+a2=2. Then the value of a1+a6+a7 is 
  1. 8
  2. 5
  3. 7
  4. 9


Solution

The correct option is C 7
Let a be the first term and d be the common difference of A.P.
Given that
a3+a5+a8=11a+2d+a+4d+a+7d=113a+13d=11     (1)

and a4+a2=2
a+3d+a+d=22a+4d=2a=12d     (2)
Putting values of a from eq(2) in eq(1), we get
3(12d)+13d=117d=14d=2
a=5
Hence, a1+a6+a7=3a+11d=7

 

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