From arithmetic progression: A2 = a1+D =10+D3 = a1+2 * D =10+2D.

In terms of geometric series: (2 * A2+2) * (2 * A2+2) = a1* 5 * A3.

Substitute above to get (20+2d) * (20+2d) =10 * 5 * (10+2d), and simplify to get d*d-3d-4=0, and get d=4 or d=- 1.

When d= 4, an = a1+(n-1) d =10+(n-1) * 4 = n+6 (n =1,2,3 ...

When d=- 1, an = a1+(n-1) d =1-n (n =1,2, 3 ...)

(2): When d is less than 0, there are (1) known an = a1+(n-1) d =1-n (n =1,2,2. ...

When n is less thAn or equal to 1 1, an is greater than or equal to 0, and when n is greater than 1 1, an is less than 0.

So,

When n is less than or equal to 1 1, the original formula =10+9+8+...+(11-n) = n * (21-n)/2.

When n is greater than 1 1,

The original formula =10+9+8+7+6+5+4+3+2+1+2+...+(n-11).

= 55+( 1+n- 1 1)/2 *(n- 1 1)

= 1 10+n *(n-2 1)/2(n = 12， 13， 14......)