两等差数列{an}、{bn}的前n项和的比Sn/S’n=（5n+3）/（2n+7）求a5/b5的值_其他解答

两等差数列{an}、{bn}的前n项和的比Sn/S’n=（5n+3）/（2n+7）求a5/b5的值

人气：236 ℃　时间：2020-03-26 04:19:45

解答

a5/b5
=(a5+a5)/(b5+b5)
=(a1+a9)/(b1+b9)
=[(a1+a9)×9÷2]/[(b1+b9)×9÷2]
=S9/S'9
=(5×9+3)/(2×9+7)
=48/25