Question 1133674
<br>
{{{(a+ar+ar^2)/(a+ar+ar^2+ar^3+ar^4+ar^5) = 27/19}}}<br>
The 4th, 5th, and 6th terms of the GP are each r^3 times the 1st, 2nd, and 3rd terms:<br>
{{{(a+ar+ar^2)/((a+ar+ar^2)(1+r^3)) = 27/19}}}<br>
{{{1/(1+r^3) = 27/19}}}<br>
{{{27(1+r^3) = 19}}}<br>
{{{27+27r^3 = 19}}}<br>
{{{27r^3 = -8}}}<br>
{{{r = -2/3}}}