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: Hint: If your series is geometric with common ratio $r$ and first value $a$... then the terms of the sequence look like $a, ar, ar^{2}, ar^{3},...$.  Write out two equations in those two unknowns ($a$ and $r$) and solve.
The equations come out to be $a+ar+ar^2=38$ and $ar+ar^3=17+1/3$.
A: $$a(1+r+r^2) = 38\\ar+ar^3 = \frac{52}3 $$ 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 $$r = \frac23, a = 18.  $$
A: TravisJ gave you the answer using definitions. The equations being $$a+ar+ar^2=38$$ $$ar+ar^3=17+\frac13=\frac{52}3$$ write the ratio to get $$\frac{a+ar+ar^2 }{ar+ar^3}=\frac{1+r+r^2 }{r+r^3}=\frac{57}{26}$$ Reduce to same denominator, simplify and you arrive to $$f(r)=57 r^3-26 r^2+31 r-26=0$$ which is a cubic equation which can be solved using radicals by Cardano method. But, without using it, what we can notice is that $$f'(r)=171 r^2-52 r+31$$ does not show any real root (negative discriminant); so, $f(r)$ is an increasing function.
By inspection, $f(0)=-26$ and $f(1)=36$; so the root is $0<r <1$. Inspection could be refined since $f(\frac12)=-\frac{79}8$ and $f(\frac34)=\frac{427}{64}$; so the root is $\frac12<r <\frac34$.
At this point, you could use Newton method starting in the middle of the last interval $x_0=\frac58$; the first iterate ise $$x_1=\frac{3727}{5572}\approx 0.66888$$ and the next is $x_2\approx 0.666673$; the solution looks to be $\frac23$ and it is easy to check that this is the solution.
Knowing $r$, use any of the two initial equations to get $a$.
