Three consecutive terms of a geometric progression are given as n-2, n and n+3. Find the common ratio
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option D
Explanation:
\(r=\frac{n}{n-2}\hspace{1mm}and\hspace{1mm}r=\frac{n+3}{n}\\∴\frac{n}{n-2}=\frac{n+3}{n}\\n^2 = n^2 +3n - 2n-6\\0=n-6\\∴n=6\\But\hspace{1mm}r = \frac{n}{n-2}\\r=\frac{6}{6-2}\\\frac{6}{4}=\frac{3}{2}\)
\(r=\frac{n}{n-2}\hspace{1mm}and\hspace{1mm}r=\frac{n+3}{n}\\∴\frac{n}{n-2}=\frac{n+3}{n}\\n^2 = n^2 +3n - 2n-6\\0=n-6\\∴n=6\\But\hspace{1mm}r = \frac{n}{n-2}\\r=\frac{6}{6-2}\\\frac{6}{4}=\frac{3}{2}\)