等差数列{an}的各项均为正数,若a3a5+a3a8+a5a10+a8a10=64,则S12=?

问题描述:

等差数列{an}的各项均为正数,若a3a5+a3a8+a5a10+a8a10=64,则S12=?

(a3a5+a3a8)+(a5a10+a8a10)
=a3(a5+a8)+a10(a5+a8)
=(a3+a10)(a5+a8)
=64
因{an}为等差数列
故a3+a10=a5+a8
故(a5+a8)^2=64
又因an>0
故a5+a8=8
S12=(a1+a12)+(a2+a11)+...+(a5+a8)+(a6+a7)
=6(a5+a8)
=48
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