Question 1209803
<br>
{{{a+ar+ar^2}}}+...<br>
That sum is 9<br>
[1] {{{a/(1-r)=9}}}<br>
{{{a^3+a^3r^3+a^3r^6}}}+...<br>
That sum is 36<br>
[2] {{{a^3/(1-r^3)=36}}}<br>
From [1],<br>
[3] {{{a^3=729(1-r)^3}}}<br>
From [2],<br>
[4] {{{a^3=36(1-r^3)}}}<br>
Equating [3] and [4]...<br>
{{{729(1-r)^3=36(1-r^3)}}}
{{{(1-r^3)/(1-r)^3=729/36=81/4}}}<br>
Simplify using {{{1-r^3=(1-r)(1+r+r^2)}}}<br>
{{{((1-r)(1+r+r^2))/(1-r)^3=81/4}}}
{{{(1+r+r^2)/(1-r)^2=81/4}}}
{{{81(1-2r+r^2)=4(1+r+r^2)}}}
{{{81-162r+81r^2=4+4r+4r^2}}}
{{{77r^2-166r+77=0}}}<br>
Use the quadratic formula to solve; choose the value of r that is less than 1, since both series converge.<br>
{{{r=(166-sqrt(166^2-4(77)(77)))/154}}}<br>
ANSWER: r = 0.6755342 to several decimal places<br>