题目

等差数列,已知等差数列｛an｝a4=6,a6=10.(1)求数列｛an｝的通项公式;
已知等差数列｛an｝a4=6,a6=10.(1)求数列｛an｝的通项公式;设等比数列｛bn｝各项均为正数,其n项和Tn,若b3=a3,T2=3,求Tn
谁把那个我解下这题,速度,谢谢

提问时间：2020-12-29

答案

等差数列通项公式an = a1 + (n-1)d
a4 = a1 + 3d =6
a6 = a1 + 5d =10
解得 d = 2,a1 = 0
所以an = 2(n-1) = 2n-2
等比数列通项公式 bn = b1*q^(n-1)
b3 = b1*q^2 = a3 = 4
T2 = b1 + b2 = b1 + b2 = b1 + b1*q =3
解方程组：
b1*q^2 =4
b1 + b1*q =3
得q=2（q为正数）
b1=1
bn = 2^(n-1)