Question

这道等比数列题该怎么做？？还有第二小问

在公比不为1的等比数列中,已知a1=2,a3,a2,a4构成等差数列,求等比数列an的通项公式及前5项和s5

Answer 1

见下图：

Answer 2

(13)

an=2.q^(n-1)
a3,a2,a4 成等差数列
a3+a4=2a2
q^2+q^3 =2q
q(q^2+q-2)=0
q(q+2)(q-1)=0
q=-2
an = -2^n
S5
= 2[ 1- (-2)^4 ]/(1+2)
=-30/3
=-10
(2)
bn= -1+(n-1)d
b2, b3-1,b4 成等比数列
b2.b4 =(b3-1)^2
(-1+d)(-1+3d) =(-2+2d)^2
3d^2-4d+1 = 4d^2 -8d+4
d^2-4d+3=0
(d-1)(d-3)=0
d=1 or 3
bn
= -1+(n-1) or -1+3(n-1)
=n-2 or 3n-4
T10
= 35 or 125