Question

已知等差数列an满足a1+a2+a3=15 a8+a9=4a4求数列通项公式

Answer 1

问：已知等差数列an满足a1+a2+a3=15 a8+a9=4a4求数列通项公式

答：设公差为d，则a1+a2+a3=3a1+3d=15a8+a9=a8+a8+2d=2a4+2d=4a4解得d=1，a1=4所以an=3n+1

答：同学，刚才是老师答错了，不好意思呀正确答案应该是设等差数列的公差为d，首项为a1，则可以列出方程组a1 + (a1 + d) + (a1 + 2d) = 15a1 + 3d + 2(a1 + 4d) = 4(a1 + 3d)化简得3a1 + 3d = 15 - 3(a1 + 2d)2a1 + 5d = 4a1 + 4d解得a1 = 3，d = 2所以该等差数列的通项公式为an = 3 + 2(n - 1)，即an = 2n + 1。