Solution: When the number of terms in arithmetic progression is 2n, the arithmetic progression of even ordinal terms is: a even number1= a1+d; A even n = a2n = a1+(2n-1) d; D even number =2d. Then: s even number =n(a even number 1+a even number n)/2 = n * [a1+d+a1+(2n-1) d]/2 = n * [a65438+] D odd number =2d. Then: s odd =n(a odd 1+a odd n)/2 = n * [a1+a1+(2n-2) d]/2 = n * [a1+(n-666). Obviously: s odd /S even = an/an+ 1. -.s2n-1= (a1+a2n-1) * (2n-1)/2 = {a1+[a1+(2n-2 D even number =2d. Then: s even number =(n- 1)[a even number 1+a even number (n-1)]/2 = (n-1) * [a1+d+a65438]. D odd number =2d. Then: s odd =n(a odd 1+a odd n)/2 = n * [a1+a1+(2n-2) d]/2 = n * [a1+(n-666).