Geometric Progression sums and sums of squares

Question

Sum of the first $4$ terms in GP is $30$ and the sum of their squares is $340$. Find the numbers. How do I solve this?

Answer 1

Let the GP be- $a,ar,ar^2,ar^3$. Sum will be- $\frac{a(1-r^4)}{1-r}=30$
Squares of terms- $a^2,a^2r^2,a^2r^4,a^2r^6.$ Sum will be-$\frac{a^2(1-r^8)}{1-r^2}=340$.
$$a(1+r^2)(1+r)=30$$ $$a^2(1+r^2)(1+r^4)=340$$ $$\frac{45}{(1+r^2)(1+r)^2}=\frac{17}{(1+r^4)}$$ $$45+45r^4=17(1+r^2)(1+r^2+2r)=17(1+r^2+2r+r^2+r^4+2r^3)$$ $$14+14r^4=17r^2+17r+17r^3$$ The two real solutions will be $r=2$,$r=\frac12$
So, $a=2,r=2$ or $a=16,r=\frac12$