Bài 2:
a: \(u_1=\dfrac{3}{1+1}=\dfrac{3}{2}\)
b: Đúng
c: \(u_{10}=u_1+9d=1+9\cdot2=19\)
d: \(u_{11}=u_7+4d\)
=>\(4d=5-6=-1\)
=>\(d=-\dfrac{1}{4}\)
\(u_9=u_7+2d=6+2\cdot\dfrac{-1}{4}=6-\dfrac{1}{2}=\dfrac{11}{2}\)
Đúng 1
Bình luận (0)