Sum of geometric series: sn = a1(1-q n)/(1-q).

The sum of infinite terms is 9, and the common ratio is positive.

∴ 0 < q < 1 (otherwise the sum of infinite terms will be positive infinity)

∴q^n=0

∴sn=a 1( 1-q^n)/( 1-q)=a 1( 1-0)/( 1-q)=a 1/( 1-q)=9①

The sum of the first two items is 5.

∴A 1+A 1q=5 ②

Combining ① and ②, we can get: q=2/3 or -2/3 (give up if it doesn't meet the meaning of the question), A 1=3.

So the fourth item A4=A 1*q? =3*(2/3)? =8/9

2:

Suppose n= 1:

The meaning of this question is:

A 1=9，A 1+A2= 12

∴a2= 12-a 1= 12-9=3

∴d=A2-A 1=3-9=-6

∴A3=A2+d=3-6=-3

∴ sum of the first 3n terms =A 1+A2+A3=9+3-3=9.

(If you need a rigorous solution, please ask)

3:

Three numbers form an arithmetic series.

Then the middle number (arithmetic average) A2=Sn/n=S3/3=36/3= 12.

Let A 1=x, then a3 = S3-a1-a2 = 36-12-x = 24-x.

If you add 1, 4, 43 to each term in turn, you will get a geometric series.

∴(x+ 1)/( 12+4)=( 12+4)/(24-x+43)

Solution: x=3 or 63

When x=63, x, 12 and 24-x are not arithmetic progression, so they are discarded.

∴x=3

The original number of ∴ is 3 122 1.

4:

∫ arithmetic progression's first n terms and sn = n * a1+n * (n-1) * d/2.

∴s 12= 12a 1+ 12* 1 1*d/2= 12a 1+66d= 12①

s30 = 30a 1+30 * 29 1 * d/2 = 30a 1+435d = 45②

From ① ②, the solution is: A 1=25/36, d =118.

The general term an = a1+(n-1) d = 25/36+(n-1)18.