课时作业(四)等差数列的性质一、选择题1若an是等差数列，且a1a4a745，a2a5a839，则a3a6a9()A39 B20C19.5 D332等差数列an中，a1a510，a47，则数列an的公差为()A1 B2C3 D43已知等差数列an满足a1a2a3a1010，则有()Aa1a1010 Ba2a1010，d1，故所求的四个数为2,0,2,4.9解析：(1)法一：根据等差数列的性质a2a10a4a82a6，由a2a6a101，得3a61，解得a6，a4a82a6.法二：设公差为d，根据等差数列的通项公式，得a2a6a10(a1d)(a15d)(a19d)3a115d，由题意知，3a115d1，即a15d.a4a82a110d2(a15d).(2)设公差为d，a1a32a2，a1a2a3153a2，a25.又a1a2a380，an是公差为正数的等差数列，a1a3(5d)(5d)16d3或d3(舍去)，a12a210d35，a11a12a133a12105.10解析：设方程的四个根a1，a2，a3，a4依次成等差数列，则a1a4a2a32，再设此等差数列的公差为d，则2a13d2，a1，d，a2，a31，a4，|mn|a1a4a2a3|.答案：C
