I have this system:
b1 + b2 + b3 = 195
b3 - b1 = 120
b1 + b1*q + b1*q^2 = 195
b1*q^2 - b1 = 120
I have to find $b_1$ and $q$(this is the private member or the progression) The answer of this exercise is $b_1 = 15$ and $q = 3$ also $b_1 = 125$ and $q = -\frac{7}{5}$. I guess the I have to get to quadratic equation but I can't. Any help is appreciated.
Hint:
$$b_1+b_1q+b_1q^2=195 \implies 1+q+q^2=195\frac1{b_1}\tag{1}$$
$$b_1q^2-b_1=120 \implies \frac1{b_1}=\frac{q^2-1}{120}\tag{2}$$