1.等比数列 a1+a3=10 a4+a6=四分之五,求a1及q a10及s10

1:
an为等比数列:
a1=a1;a3=a1*q^2;a4=a1*q^3;a6=a1*q^5;
a1+a3=10 ==>
a1+a1*q^2=10;==>a1(1+q^2)=10---(1);
a4+a6=5/4;===>
a1*^q3+a1*q^5=5/4==>a1*q^3*(1+q^2)=5/4----(2);
(1),(2)式左右相除;1/q^3=8;==>q=1/2;
将q=1/2代入(1)式得:a1=8;
a10=a1*q^9=8*(1/2)^9=1/64;
s10=8*[1-(1/2)^10]/(1-1/2)=1023/64;
2:
等差数列abc==>
a+c=2b;
a+b+c=12;==>3b=12==>b=4;
==>a+c=8;c=8-a;
a,b,c+2成等比数列==>a*(c+2)=b^2;==>a*(8-a+2)=4^2==>a^2-10a+16=0;
==>a=2,或a=8;
3:
等比数列a1*a2*a3=1 ;==>a1*a1q*a1q^2=a1^3q^3=1;-==>a1*q=1==>a2=1;
a2+a3+a4=7==>a2+a2q+a2q^2=7==>q+q^2=6;==>
q=-3或q=2;
当q=-3;
a1=-1/3;a5=a1*q^4=(-1/3)*(-3)^4=-27;
s5=a1+(a2+a3+a4)+a5=-1/3+7-27=-20-1/3=-61/3;
当q=2;
a1=1/2;a5=a1*q^4=1/2*2^4=8;
s5=a1+(a2+a3+a4)+a5=1/2+7+8=31/2;