Question 983219


Let &nbsp;<B>a</B>&nbsp; be the first term of our geometric progression and &nbsp;<B>r</B>&nbsp; be its common ratio. 


Then our first equation is


{{{a/(1-r)}}} = {{{4}}}.


It is the formula for the sum of infinite geometric progression.


The sequence consisting of cubes of geometric progression is geometric progression itself, &nbsp;with the first term &nbsp;{{{a^3}}}&nbsp;  and the common ratio &nbsp;{{{r^3}}}. &nbsp;It is easy to check. 


Therefore, &nbsp;the sum of such progression is 


{{{a^3/(1-r^3)}}} = {{{192}}}.


It is our second equation.


Now, &nbsp;divide the second equation by the first one. &nbsp;You will get


{{{a^2/(1+r+r^2)}}} = {{{48}}}.


Thus you decreased the degree of the second equation from &nbsp;3&nbsp; to &nbsp;2.


Next, &nbsp;express the term &nbsp;<B>a</B>&nbsp; from the first equation as 


{{{a}}} = {{{4*(1-r)}}} 


and substitute it into the third equation. &nbsp;You will get 


{{{(4^2*(1-r)^2)/(1+r+r^2)}}} = {{{48}}}, &nbsp;&nbsp;or


{{{(1-r)^2}}} = {{{3*(1+r+r^2)}}}, &nbsp;&nbsp;or


{{{1 -2r + r^2}}} = {{{3r^2 + 3r +3}}}.


It is a quadratic equation.


Would you solve it and complete the solution from this point?