I am stuck solving for $a-ar^2=112$ and $ar-ar^3=84$
I got $a=\frac{112}{1-r^2}$ and $a=\frac{84}{r-r^3}$
Then I got a cubic equation. But answer only has 1 value for a & r, so I think there must be an easier way?
I am stuck solving for $a-ar^2=112$ and $ar-ar^3=84$
I got $a=\frac{112}{1-r^2}$ and $a=\frac{84}{r-r^3}$
Then I got a cubic equation. But answer only has 1 value for a & r, so I think there must be an easier way?
$a-ar^2=112\iff a(1-r^2)=112\qquad \qquad(1)$
$ar-ar^3=84\iff ar(1-r^2)=84\qquad\qquad(2)$
Since $r=1$ and $r=-1$ are not solutions of the equations, the equation on the right in (1) is equivalent to
$ a={112\over 1-r^2}. $
Now, substituting this into the equation on the right of (2):
$ {112\over 1-r^2}\cdot r (1-r^2)=84\iff112 r =84\iff r={84\over112}={3\over4}$.
Going back to the first equation
$a(1-(3/4)^2)=112\Rightarrow a={112\over 1-{9\over16}}={112\over 7/16}={16\cdot 112\over 7}=16^2=256$.