The first kind:

Sum of the first n terms in geometric series

(1) when q= 1, Sn=n×a 1 is brought into q= 1, which does not conform to the meaning of the question (if q= 1, then a1= an is obtained from a 1+an=66.

② When q≠ 1,

sn=a 1*( 1-q^n)/( 1-q)= 126；

a 1+[a 1 *q^(n- 1]]= 66；

(a 1 * q)- 1 = 128；

Let's call it a day, starting with (a1* q)-1=128; Get a 1= 129/q and substitute it into the above two formulas.

Remove n first and then find q, find it yourself! Do more work and get better slowly!

If I can't find out, I can calculate it again!

The second type:

a2*a(n- 1)=a 1*an

A 1*an= 128 is available.

And a 1+an=66.

We can find a 1=2, an=64 or a 1=64, an=2.

1) when a 1=2, an=64 and sn= 126 are parallel, n=6 q=2 can be obtained.

2) when a 1=64, an=2 are connected in parallel, and sn= 126, n=6 q= 1/2 can be obtained.

In fact, mathematics is like this. Many things have simple algorithms, but it doesn't matter if you can't think of them. Don't be afraid, just count them directly.