已知等比数列{aN}中,各项都是正数,且a1,1⼀2(a3),2a2成等差数列,则a9+a10⼀a7+a8=

2025-03-01 11:02:47
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回答1:

由题意易知 a3 = a1 + 2a2
a1 * q^2 = a1 + 2a1 * q (a1不等于0)
即 q^2 - 2q - 1 = 0 , 解得 q = 1 + √2 或 -1 + √2 (√2 指根号2)
(a9+a10)/(a7+a8) = a9(1+q)/[a7(1+q)]
= a9/a7
= q^2
= 3 + 2√2 或 3 - 2√2