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Question

If a1,a2,a3,a4, are in A.P. and a1+a5+a9+...+a29=64, then the value of a3+a8+a22+a27 is

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Solution

a1+a5+a9+...+a29=64
(a1+a29)+(a5+a25)+(a9+a21)+(a13+a17)=64
4k=64
k=16

a3+a8+a22+a27=(a3+a27)+(a8+a22)=2k=32

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