Question

已知{an}为单调递增的等比数列，Sn为其前n项和，满足S4=a1+28,且a2,a3+2,a4仍构

已知{an}为单调递增的等比数列，Sn为其前n项和，满足S4=a1+28,且a2,a3+2,a4仍构成等差数列。(1)求通项{an}公式

Answer 1

an =a1.q^(n-1)

S4=a1+28
a1(1+q+q^2+q^3) = a1+28
a1(q+q^2+q^3)=28 (1)

a2,a3+2,a4仍构成等差数列

a2+a4= 2(a3+2)
a1(q+q^3) = 2(a1q^2 + 2)
a1(q+q^3-2q^2) =4 (2)

(1)/(2)
(q+q^2+q^3)/(q+q^3-2q^2)=7
1+q+q^2=7(1+q^2-2q)
2q^2-5q+2=0
(2q-1)(q-2)=0
q=2
a1= 2
an = 2^n

Answer 2

∵S4=a1+28,∴a2+a3+a4=28①,
又2(a3+2)=a2+a4,∴2(a3+2)+a3=28,∴a3=8②,
∵{an}等比,∴由①②可得：
a1(q+q²+q³)=28③,a1·q²=8④,
③÷④,整理得：2q²-5q+2=0,∴q=2,或q=½,
∵{an}单调递增,∴q=2,a1=2,
∴an=2∧n(n∈N*)。