Question 1171167
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The solution from the other tutor is a good example of the fact that there are practically endless ways to solve any given problem....<br>
Here is a slightly simpler method.<br>
(1) {{{a+ar = 108}}}  the sum of the 1st and 2nd terms is 108<br>
(2) {{{ar^2+ar^3 = 9}}}  the sum of the 3rd and 4th terms is 12<br>
Factor out a common factor on the left in (2); then substitute (1):<br>
{{{ar^2+ar^3 = r^2(a+ar) = 108r^2 = 12}}}
{{{r^2 = 12/108 = 1/9}}}
{{{r = 1/3}}} or {{{r = -1/3}}}<br>
If r=1/3 then<br>
{{{a+a(1/3) = 108}}}
{{{(4/3)a = 108}}}
{{{a = 108(3/4) = 81}}}<br>
ANSWER #1: a=81; r=1/3.  The terms are 81, 27, 9, 3.<br>
CHECK: 81+27 = 108; 9+3 = 12.<br>
If r=-1/3 then<br>
{{{a+a(-1/3) = 108}}}
{{{(2/3)a = 108}}}
{{{a = 108(3/2) = 162}}}<br>
ANSWER #2: a=162; r=-1/3.  The terms are 162, -54, 18, -6.<br>
CHECK: 162+(-54) = 108; 18+(-6) = 12.<br>