Sum of an infinite $G.P$ is $2020$. Each term of this $G.P$ is squared to make a new series whose sum is $20200$. If the common ratio of the original $G.P$ is ${a\over b}$ where $gcd(a,b)=1$, evaluate $b-a$.
My Work:-
Well sum of an infinte $GP$ = ${a\over 1-r}$ where $a$ is the first term and $r$ is the common ratio. Also we can simply observe that whenever we square the whole $G.P$ ; $a$ changes to $a^2$ and $r$ changes to $r^2$ as well. Thus basically we have $2$ equations :-
$a = 2020 + 2020r$
$a^2 = 20200 + 20200r^2$
Now I am stuck as to how to solve these equations.
$NOTE:-$ $G.P$ $STANDS$ $FOR$ $GEOMETRIC$ $PROGRESSION$
An alternative that avoids the ratio arithmetic is to write the series as $$ s \ = \ c \ + \ c·\left(\frac{a}{b} \right) \ + \ c·\left(\frac{a}{b} \right)^2 \ + \ c·\left(\frac{a}{b} \right)^3 \ + \ \cdots $$ and $$ 10·s \ = \ c^2 \ + \ c^2·\left(\frac{a}{b} \right)^2 \ + \ c^2·\left(\frac{a}{b} \right)^4 \ + \ c^2·\left(\frac{a}{b} \right)^6 \ + \ \cdots \ \ , $$ with $ \ c \ $ being the initial term since $ \ a \ $ is already "taken". We can produce a third series by multiplying the first one by $ \ c \ $ to obtain $$ c·s \ = \ c^2 \ + \ c^2·\left(\frac{a}{b} \right) \ + \ c^2·\left(\frac{a}{b} \right)^2 \ + \ c^2·\left(\frac{a}{b} \right)^3 \ + \ \cdots $$
Subtracting the second series from the third leads to $$ c·s \ - \ 10·s \ = \ c^2·\left(\frac{a}{b} \right) \ + \ c^2·\left(\frac{a}{b} \right)^3 \ + \ c^2·\left(\frac{a}{b} \right)^5 \ + \ \cdots \ = \ \left(\frac{a}{b} \right) · 10 · s $$ $$ \Rightarrow \ \ c \ = \ 10 · \left[ \ \left(\frac{a}{b} \right) + 1 \ \right] \ \ . $$
The sum for the first infinite series is given by $ \ s \ = \ \frac{c}{1 - \left(\frac{a}{b} \right)} \ \Rightarrow \ c \ = \ s · \left[ \ 1 - \left(\frac{a}{b} \right) \ \right] \ \ , $ so with $ \ s \ = \ 2020 \ , $ we find $$ 10 · \left[ \ \left(\frac{a}{b} \right) + 1 \ \right] \ = \ 2020 · \left[ \ 1 - \left(\frac{a}{b} \right) \ \right] \ \ \Rightarrow \ \ 2030 · \left(\frac{a}{b} \right) \ = \ 2010 $$ $$ \Rightarrow \ \ \left(\frac{a}{b} \right) \ = \ \frac{201}{203} \ \ , \ \ b - a \ = \ 2 \ \ , $$ since $ \ \gcd(201,203) = 1 \ \ . $ We can also determine the initial term to be $$ \ c \ = \ 10 · \left(\frac{201 \ + \ 203}{203} \right) \ = \ \frac{4040}{203} \ \approx \ 19.901 \ \ . $$
This can then be generalized: if the ratio of the sum of the "squared-term" series to the sum $ \ s \ $ of the original series is $ \ n \ \ , $ then $$ \ \frac{a}{b} \ = \ \frac{s-n}{s+n} \ \ \ \text{and} \ \ \ c \ = \ \frac{2·n·s}{s+n} \ \ . $$
(The difference $ \ b - a \ $ would need to be found after $ \ \frac{a}{b} \ $ is reduced to "lowest terms".)