在等差数列｛an｝中a1大于0,a2a4＋2a3a5＋a4a6＝25求a3＋a5的值
人气：211 ℃ 时间：2020-06-05 04:09:18
解答
a2a4+2a3a5+a4a6=(a3-d)(a3+d)+2a3a5+(a5-d)(a5+d)=(a3)^2-d^2+2a3a5+(a5)^2-d^2=(a3)^2+2a3a5+(a5)^2-2d^2=(a3+a5)^2-2d^2=25 因an＞0,a1＞0,d＞0(a3+a5)^2-2d^2=25 a3+a5=√(25+2d^2)
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