Question

设{an}是公比大于1的等比数列,若a1+a2+a3=7,且a1+3,3a2,a3+4构成等比数列,求数列{an}的通项公式

急求，师哥师姐快来帮忙

Answer 1

好像无实根啊，题错了？
a1+a2+a3=7, a2=7- a1- a3, a22=a12+a32+49+a1a3-7a1-7a3
a1xa3=a22 = a12+a32+49+a1a3-7a1-7a3, a12+a32-7a1-7a3+49=0 ------- ①
(a1+3)(a3+4)=(3a2)2= 9a22=9a1a3, 8a1a3-4a1=3a3+12, a1(8a3-4)=3a3+12,
a1=(3a3+12)/(8a3-4) ------ ②

a1+a3=x, a1a3=y=a22, a2=7-x, y=a22=x2-14x+49
a12+2a1a3+a32-7a1-7a3+49=2a1a3, (a1+a3)2-7(a1+a3)+49=2a1a3, x2-7x+49=2y
x2-7x+49=2(x2-14x+49)=2x2-28x+98, x2-21x+49=0, x=(21+-7√5)/2

x=(21-7√5)/2 = 2.674
y=[(21-7√5)/2]2-14*(21-7√5)/2+49
=(441-294√5+245)/4-147+49√5+49
=(147-49√5)/2 = 18.716
a1=x-a3=2.674-a3
y=a1a3=a3(2.674-a3)= 18.716
a32-2.674a3+18.716=0 (无实根)

x=(21+7√5)/2 = 18.326
y=[(21+7√5)/2]2-14*(21+7√5)/2+49
=(441+294√5+245)/4-147-49√5+49
=(147+49√5)/2 = 128.284
a1=x-a3=18.326-a3
y=a1a3=a3(18.326-a3)= 128.284
a32-18.326a3+128.284=0 (无实根)

Answer 2

设公比为q,由a1+a2+a3=7得a1+a1q+a1q^2=7，又3a2是等差中项，所以2*（3a1q）=a1+3+a1q^2+4,联立两式解方程组，解得q=2，a1=1,所以an=a1q^(n-1)=2^(n-1)

Answer 3

题错了