China Naming Network - Eight-character lottery - Let the sum of the first n terms of arithmetic progression {an} be Sn; if S3=9 and S6=36, then A7+A8+A9 = _ _ _ _ _.

Let the sum of the first n terms of arithmetic progression {an} be Sn; if S3=9 and S6=36, then A7+A8+A9 = _ _ _ _ _.

Thinking of solving the problem: subtract S 3 from S 6 to get the value of a 4+a 5+a 6, then use the properties of arithmetic progression to find the sum of a 4+a 5+a 6 and a 1+a 2+a 3, that is, the relationship with S 3, and then use the properties of arithmetic progression to find the relationship between a 7+a 8+a 9 and D and S 3.

a4+a5+a6=S6-S3=36-9=27，

a4+a5+a6 =(a 1+3d)+(a2+3d)+(a3+3d)=(a 1+a2+a3)+9d = S3+9d = 9+9d = 27、

So d=2,

Then a7+A8+A9 = (a1+6d)+(a2+6d)+(a3+6d) = S3+18d = 9+36 = 45.

So the answer is: 45.

,4,

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