Problem description:
1. In the geometric series {an} with the first term of 3, if the nth term is 48 and the 2n-3 term is 192, how to find n?
2. In the geometric series {an} whose common ratio is an integer, if a 1+a4 = 18 and a2+a3 = 12, what is the sum of the first eight terms of this series?
13. In the geometric series {an}, what are the terms of Sn = 93, an = 48, q = 2, and n?
Analysis:
1:3*x^(n- 1)=48； 3*x^(2n-3- 1)= 192
= & gtx^(n-3)=4； x^(n- 1)= 16
= & gtx^2=4=>； X=2 or x=-2.
=>3 * 2 (n-1) = 48 or 3 * (-2) (n-1) = 48?
= & gtn=5
2:a 1+a4=a 1+a 1*x^3=a 1( 1+x^3)= 18； a2+a3=a 1(x+x^2)= 12
= & gt(x+x^3)/(x+x^2)= 18/ 12=3/2
= & gt( 1+x)x/( 1+x)( 1-x+x^2)=3/2
=> 1+x=0 or x/(1-x+x 2) = 3/2.
= & gtx=- 1 or x=2 or x= 1/2
=>x is an integer = & gtx=- 1 or x=2.
=> if x=- 1 then a1(1+(-1) 3) =18 = > This is wrong.
=> if x=2, then a1(1+2 3) =18 = > a 1=2
= & gta 1+a2+...+a8=2*( 1-2^8)/( 1-2)=5 10
3:sn=a 1*( 1-x^n)/( 1-x)=>； 93=a 1*(2^n- 1)； an=48= >a 1*2^(n- 1)=48
= & gta 1*x^n=96=>； a 1=96/2^n
= & gt93=96*(2^n- 1)/2^n=96-96/2^n
= & gt96/2^n=3=>； 2^n=32=>； n=5