Question 941784
The formula for sum of infinite GP is {{{ a/(1-r)}}}
 hence {{{ a/(1-r) =15}}}..........eq(1)
   
HE SUM OF THE SQUARES OF THESE TERMS IS 45
  hence {{{ a^2/(1-r^2) =45}}} ......eq(2)
  divide eq(1) with eq (2)
  {{{ ( a/(1-r))/((a^2/(1-r^2))) =15/45}}}
  {{{ ( a/(1-r))/((a^2/(1-r)(1+r)))) =15/45}}}
    {{{(1/1)/(a/(1+r) )=1/3}}}
     {{{(1+r)/a=1/3}}}
 {{{a/(1+r) =3}}}
    move (1+r) to the right
     {{{ a= (1+r)*3}}}
 but from eq(1) {{{ a/(1-r) =15}}}
             move (1-r) to the right
           {{{ a= (1-r)*15}}}
            from above expressions
        {{{(1+r)*3  = (1-r)*15}}}
         3+3*r= 15*1-r*15
       3+3r =15-15r
        move -15r to the left
      3+3r+15r =15
       3+18r =15
         18r =15-3
      18r =12
         r = {{{ 12/18}}}
           r= {{{2/3}}}
   but a  = (1+r)*3
           = {{{(1+(2/3))*3}}}
                ={{{((3/3) +(2/3))*3}}}
 = {{{ ((3+2)/3 )*3}}}
 = 5
a= 5, r= 2/3
series  =  5, 10/3, 20/9,...