a 19=a9*q^ 10

a20=a 10*q^ 10

Addition of two formulas

a 19+a20=(a9+a 10)*q^ 10

Namely: 4 = 2 * q 10

q^ 10=2

In the same way.

a29+a30=(a 19+a20)*q^ 10=4*2=8

(2)

Multiply A5 = A 1 * 3 4

Get 162=8 1*a 1.

That is a 1=2.

By sn = (q (n+1)-1) * a1/(q-1)

Get 242 = (3 (n+1)-1) * 2/2.

That is 3 (n+ 1) = 243.

n+ 1=5

n=4

(3)

Multiply a5=a3+2d

Get 9=3+2d

That is, the tolerance d=3.

{an}={-3，0，3，6，9，...}

{bn} is a geometric series.

The first item B 1 = 3-3 = 1/27.

Q = 3 3 = 27。

The sum of the first n items is (27 (n+ 1)- 1)/702.

(4)

Let these three numbers be x-d, x, x+d.

Then the sum of three numbers 3x= 12.

x=4

And 4-d, 4, 13+d is a geometric series.

So 4*4=(4-d)*( 13+d)

That is, d=3.

These three numbers are 1, 4, 7.

(5)

Let the first term be a and the common ratio be q.

Then A4 = AQ 3 and A7 = AQ 6.

It is known that a4*a7=-5 12.

That is, a 2q 9 =-2 9.

And q is an integer.

There must be q=-2 and a 2 =1.

It is also known that a3+a8= 124.

That is -4a+ 128a= 124.

a= 1

So a 10 = AQ 9 =-5 12.