Math about Geometric series In a geometric series, the sum of $1^{st}$ term $+$ $2^{nd}$ term $+$ $3^{rd}$ term $= 38$, 
the sum of $2^{nd}$ term $+ 4^{th}$ term $=  17 \frac{1}{3}$; 
how to calculate the common ratio? ( it is for sure the answer is $r= \frac{2}{3}$ and the first term $= 18$)

$$a(1+r+r^2) = 38\\ar+ar^3 = \frac{52}3 $$ eliminating $a$ leaves us with 
$$\frac{1+r+r^2}{r+r^3} = \frac{57}{26}.$$ this can be rewritten as $$ 0=57r^3-26r^2+31r-26 = (3r-2)(19r^2+4r+13).$$ the quadratic equation $19r^2 + 4r+13=0 $ has no real roots, therefore $$r = \frac23, a = 18.  $$
