等差数列{an}中,3a5=7a10,且a1

问题描述:

等差数列{an}中,3a5=7a10,且a1

a5=a1+4d,a10=a1+9d
那么由3a5=7a10,得:
3(a1+4d)=7(a1+9d)
所以d=-4a1/51
那么an=a1+(n-1)d
=a1-4(n-1)a1/51
=(55-4n)a1/51
令an>0,则55-4n55,n>55/4
则a14>0,a13