Three consecutive terms of a geometric progression are give as n - 2, n and n + 3. Find the common ratio
\(\frac{3}{2}\)
\(\frac{2}{3}\)
\(\frac{1}{2}\)
\(\frac{1}{4}\)
Correct answer is A
\(\frac{h}{n - 2} = \frac{n + 3}{n}\)
n\(^2\) = (n + 3) (n - 2)
n\(^2\) = n\(^2\) + n - 6
n\(^2\) + n - 6 - n\(^2\) = 0
n - 6 = 0
n = 6
Common ratio: \(\frac{n}{n - 2} = \frac{6}{6 - 2} = \frac{6}{4}\) = \(\frac{3}{2}\)