Let n =0, A22 = 3a1+(n-1) d = 60-3 (n-1) = 0, n=2 1, that is, in the original series, 2 1 item.

Divide the first 30 items of the new series into two series-the absolute value of the first item to 2 1 item of the original series; Absolute values of items 22 and 30.

The sum of the first part is sn1= (a1+an) n/2 = (60+0) * 21/2 = 30 * 21= 630.

Think of the second part as another series (that is, a22 is a 1 and a30 is a9), and A9 = a1+(n-1) d = 3+(9-1) * 3 = 27.

sn2 =(a 1+an)n/2 =(3+27)* 9/2 = 135

The sum of the top 30 items in the new series sn = sn1+sn2 = 630+135 = 765.