Question

Sum of infinity question please help!

Answer 1

`a_2=6;\ S_n=24\\\\S_n=(a_1)/(1-r);\ a_1=(a_2)/(r)\\\\therefore\ S_n=((a_2)/(r))/(1-r)=(a_2)/(r(1-r))=(a_2)/(r-r^2)\\\\subtitute:\\\\(6)/(r-r^2)=24\\\\(6)/(r-r^2)=(24)/(1)\ \ \ \ \ |cross\ multiply\\\\24(r-r^2)=1\cdot6\\\\24r-24r^2=6\ \ \ \ |divide\ both\ sides\ by\ 6\\\\4r-4r^2=1\ \ \ \ \ |subtract\ 1\ from\ both\ sides\\\\4r-4r^2-1=0\\\\-4r^2+4r-1=0\ \ \ \ \ |change\ signs\\\\4r^2-4r+1=0\\\\(2r)^2-2\cdot2r\cdot1+1^2=0\ \ \ \ |use\ a^2-2ab+b^2=(a-b)^2`


`(2r-1)^2=0\iff2r-1=0\ \ \ \ \ |add\ 1\ to\ both\ sides\\\\2r=1\ \ \ \ \ \ |divide\ both\ sides\ by\ 2\\\\\boxed{r=(1)/(2)}`