设Sn是等差数列an的前n项和,且(1/3)S3与(1/4)S4的等比中项为(1/5)S5,(1/3)S3与(1/4)S4的等差中项

问题描述:

设Sn是等差数列an的前n项和,且(1/3)S3与(1/4)S4的等比中项为(1/5)S5,(1/3)S3与(1/4)S4的等差中项
设Sn是等差数列an的前n项和,且(1/3)S3与(1/4)S4的等比中项为(1/5)S5,(1/3)S3与(1/4)S4的等差中项为1,求等差数列的通项公式an

设an=a1+(n-1)d S3=a1+a2+a3(1/3)S3=a1+d
(1/4)S4=a1+(3\2)d
(1\5)S5=a1+2d
(1/3)S3+(1/4)S4=2a1+5\2d=2(1/3)S3*(1/4)S4=(1\5)S5*(1\5)
解得a1=4\7 d=12\35 所以an=12\35n+8\35