Infinite geometric progression involving square terms The sum of an infinite geometric progression is 15 and the sum the squares of these terms is 45. Find the series. 
The formula for sum of infinite GP is $\frac{a }{1-r} $ and I got two equations $15=\frac {a}{1-r} $ snd $45=\frac {a^2}{1-r^2} $. Solving I got $3=\frac{a}{1+r} $

Clearly, $a\ne0, r^2\ne1$
We have $\displaystyle\frac{\dfrac{a^2}{1-r^2}}{\left(\dfrac a{1-r}\right)^2}=\frac{1-r}{1+r}$
But, $\displaystyle\frac{\dfrac{a^2}{1-r^2}}{\left(\dfrac a{1-r}\right)^2}=\frac{45}{15^2}=\frac15$
Now apply Componendo and dividendo
