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
A
−8
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B
5
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C
7
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D
9
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Solution
The correct option is C7 Let a be the first term and d be the common difference of A.P. Given that a3+a5+a8=11⇒a+2d+a+4d+a+7d=11⇒3a+13d=11⋯(1)
and a4+a2=−2 ⇒a+3d+a+d=−2⇒2a+4d=−2⇒a=−1−2d⋯(2) Putting values of a from eq(2) in eq(1), we get 3(−1−2d)+13d=11⇒7d=14⇒d=2 ∴a=−5 Hence, a1+a6+a7=3a+11d=7