据前n项和公式Sn=na1+n(n-1)*d/2S2n=2na1+2n(2n-1)*d/2将两式分别化成以n为根的二次项方程得Sn=(d/2)n^2+(a1-d/2)nS2n=(2d)n^2+(2a1-d)nSn/S2n为常数,即以上两方程根系数（-b/2a）相同列式化简得(2a1-d)d=(2a1-d)*2d...