Solution: a 1, a2, a3, a4, a5, a6, a7 are known.

(a 1+a3+a5+a7)-(a2+a4+a6)= 42( 1)，a 1+a4+a7=27 (2)

And a 1, a3, a5 and a7 are arithmetic progression, then a 1+a7=a3+a5.

A2, A4, A6 are geometric series, then there is Q that makes a4=a2*q a6=a2*q? (3)

It is easy to know that A 1, A2, A3, A4, A5, A6 and A7 are integers.

From (1)(2), a1+a3+a5+a7 = 2 (a1+a7) = 2 [27-a4] = 42+(a2+a4+a6).

According to (3)

a2(q？ +3q+ 1)= 12

12 has a factor of 1, 2, 3, 4, 6, 12.

Q is obviously an integer.

Then q can only be 1.

q？ +3q=0 a2= 12

Q takes 0 or -3.

So a4=0 or a4=-36.