Question 1112775
.
<pre>
We are given  {{{a + ar}}} = 90,   or  {{{a*(1+r)}}} = 90,   or  a = {{{90/(1+r)}}}.


We are also given that  {{{a/(1-r)}}} = {{{640/7}}},   or   a = {{{(640/7)*(1-r)}}}.


Since left sides of the two last equations are identical, their right sides are equal:

{{{90/(1+r)}}} = {{{(640/7)*(1-r)}}},


which implies

90*7 = {{{640*(1-r^2)}}},

63 = {{{64 - 64*r^2}}}   ====>  {{{64r^2}}} = 64 - 63 = 1  ====>  {{{r^2}}} = {{{1/64}}}  ====>  r = +/- {{{sqrt(1/64)}}} = +/- {{{1/8}}}.


<U>Answer</U>.  The two value of the common ratio are  +/-{{{1/8}}}.
</pre>

Solved.