An algebra problem by edil tizon

Pure algebra. Put $b = qa, c = q^2a$ . Then we have $a, 2qa, 3q^2a$ are in arithmetic progression so we have $2qa-a = 3q^2a - 2qa$ Note $a \not = 0$ couse the geometric progression $a,b,c$ is non-costant, so we can divide by $a$ and get $2q-1 = 3q^2-2q$ this means $3q^2 - 4q +1 = 0$ easily factorizable in $(3q -1)(q-1) = 0$ The solution $q=1$ is not acceptable because, again, $a,b,c$ is non-constant. So in the end we have $q = \dfrac{1}{3}$ .

From what is given, we know that b2=ac and 4b=3c+a. If we square the last equality, we get 16b2=a2+6ac+9c2. Substituting b2 for ac, the last becomes a2−10ac+9c2=0, which means that either a=c or a=9c. Let us assume that a=c. Then the arithmetic progression becomes c,2b,3c, which yields 4b=3c+c=4c, or b=c, which is not a solution (since the geometric progression is not constant). Then a=9c. The arithmetic progression becomes 9c,2b,3c, which means that 4b=9c+3c=12c, or b=3c. The geometric progression becomes 9c,3c,c, which obviously has a quotient q=13.