Question

How many geometric progressions is/are possible containing 27, 8 and 12 as three of its/their terms?

• A. One
• B. Two
• C. Four
• D. Infinitely many

Answer 1

The correct option is D Infinitely many

Let $r_1=\frac{12}{8}=\frac{3}{2}$ and $r_2=\frac{27}{12}=\frac{9}{4}=r_1^2$

As long as this ratio between $r_1$ and $r_2$ is maintained and any ${}^{\prime }{n}^{\prime }$ number of terms are added between $8$ and $12$ and $(n+1{)}^2$ number of terms are added between $12$ and $27$, we have a GP. Hence, there are infinitely many terms added.