To find the common ratio of the geometric progression with terms \( n-2 \), \( n \), and \( n+3 \), we use the property that the square of the middle term is equal to the product of the other two terms: \(n^2 = (n - 2)(n + 3)\) Expanding the right side gives: \(n^2 = n^2 + 3n - 2n - 6\) Simplifying leads to: \(0 = n - 6 \implies n = 6\) The terms of the progression are: - First term: \( n - 2 = 4 \) - Second term: \( n = 6 \) - Third term: \( n + 3 = 9 \) The common ratio \( r \) is calculated as follows: \(r = \frac{n}{n-2} = \frac{6}{4} = \frac{3}{2}\) We can verify it with the third term: \(r = \frac{n+3}{n} = \frac{9}{6} = \frac{3}{2}\) Thus, the common ratio of the geometric progression is: \(\boxed{\frac{3}{2}}\)