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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.

Matthew Cassell
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user202029
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1 Answers1

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Hint:

$$b_1+b_1q+b_1q^2=195 \implies 1+q+q^2=195\frac1{b_1}\tag{1}$$

$$b_1q^2-b_1=120 \implies \frac1{b_1}=\frac{q^2-1}{120}\tag{2}$$

Daniel R
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