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Question

For an A.P., a1, a2, a3,, if a3+a5+a8=11 and a4+a2=2, then the value of a1+a6+a7 is

A
17
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B
13
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C
3
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D
7
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Solution

The correct option is D 7
Let first term =a and common difference =d
a3+a5+a8=11
(a+2d)+(a+4d)+(a+7d)=11
3a+13d=11 (1)

a4+a2=2
(a+3d)+(a+d)=2
a+2d=1 (2)

Solving equation (1) and (2), we get
a=5 , d=2
Now, a1+a6+a7
=3a+11d
=15+22=7

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