Question 1131206


so formula for the sum of infinite geometric progression: {{{sum=a[1]/(1-r)}}}

given:

 {{{sum=36/7}}} 
{{{r=-2/3}}}


{{{36/7=a[1]/(1-(-2/3))}}}


{{{36/7=a[1]/(1+2/3)}}}


{{{36/7=a[1]/(5/3)}}}


{{{(36/7)(5/3)=a[1]}}}


{{{(cross(36)12/7)(5/cross(3))=a[1]}}}


{{{(12/7)(5/1)=a[1]}}}


{{{a[1]=(60/7)}}}