Three consecutive terms of a geometric progression are given as n-2, n and n=3. Find the common ratio

In a geometric progression, each term is obtained by multiplying the previous term by a fixed constant called the common ratio (r). Therefore, if we divide any two consecutive terms in a geometric progression, we get the same value for the common ratio. In this problem, the three consecutive terms are given as n-2, n, and n+3. We can divide the second term (n) by the first term (n-2) to get: n / (n-2) We can also divide the third term (n+3) by the second term (n) to get: (n+3) / n Since these two values must be equal to the common ratio (r), we can set them equal to each other and solve for r: n / (n-2) = (n+3) / n n^2 = n(n-2) + (n+3)(n-2) n^2 = n^2 - 2n + n^2 - 5n + 6 0 = n^2 - 7n + 6 0 = (n-1)(n-6) This equation has two solutions, n=1 and n=6. However, n cannot be equal to 1 because that would make the first term (n-2) equal to -1, which is not allowed in a geometric progression. Therefore, n must be equal to 6. Plugging n=6 back into the equation for the common ratio, we get: r = (n+3) / n = 9 / 6 = 3/2 Therefore, the correct answer is: 3/2.