关于等比数列的3道数学题,
问题描述：
关于等比数列的3道数学题,
⒈在首项为3的等比数列{an}中,若第n项是48,第2n - 3项是192,求n?
⒉在公比为整数的等比数列{an}中,如果a1 + a4 = 18,a2 + a3 = 12,那么该数列的前8项之和是多少?
⒊在等比数列{an}中,Sn = 93,an = 48,公比q = 2,则项数n等于多少?
答
1:3*x^(n-1)=48;3*x^(2n-3-1)=192
=>x^(n-3)=4;x^(n-1)=16
=>x^2=4=>x=2 or x=-2
=>3*2^(n-1)=48 or 3*(-2)^(n-1)=48
=>n=5
2:a1+a4=a1+a1*x^3=a1(1+x^3)=18;a2+a3=a1(x+x^2)=12
=>(x+x^3)/(x+x^2)=18/12=3/2
=>(1+x)x/(1+x)(1-x+x^2)=3/2
=>1+x=0 or x/(1-x+x^2)=3/2
=>x=-1 or x=2 or x=1/2
=>x is integer=>x=-1 or x=2
=>if x=-1 then a1(1+(-1)^3)=18=>It is wrong.
=>if x=2 then a1(1+2^3)=18=>a1=2
=>a1+a2+...+a8=2*(1-2^8)/(1-2)=510
3:Sn=a1*(1-x^n)/(1-x)=>93=a1*(2^n-1);an=48=>a1*2^(n-1)=48
=>a1*x^n=96=>a1=96/2^n
=>93=96*(2^n-1)/2^n=96-96/2^n
=>96/2^n=3=>2^n=32=>n=5