Let's denote the first term of the geometric progression (GP) as \( a \) and the common ratio as \( r \).
According to the problem, we have:
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The sum of the first two terms: \[ a + ar = 20 \] This can be factored to: \[ a(1 + r) = 20 \quad \text{(1)} \]
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The sum of the second and third terms: \[ ar + ar^2 = 30 \] This can be factored to: \[ ar(1 + r) = 30 \quad \text{(2)} \]
Now, let's divide equation (2) by equation (1) to eliminate \( a \):
\[ \frac{ar(1 + r)}{a(1 + r)} = \frac{30}{20} \]
Provided \( 1 + r \neq 0 \): \[ r = \frac{3}{2} \]
Now that we have the common ratio \( r \), we will substitute it back into equation (1) to find \( a \):
Substituting \( r = \frac{3}{2} \) into equation (1): \[ a(1 + \frac{3}{2}) = 20 \] This simplifies to: \[ a \left(\frac{5}{2}\right) = 20 \] Multiplying both sides by \( \frac{2}{5} \): \[ a = 20 \cdot \frac{2}{5} = 8 \]
Thus, we have:
- First term \( a = 8 \)
- Common ratio \( r = \frac{3}{2} \)
So, the common ratio is \( \frac{3}{2} \).