Q. Which term of the GP $3,\ 3\sqrt{3},9\ldots$ is 2187?
A) 15
B) 14
C) 13
D) 19
Explanation
(c)
Let $n$th term of GP is 2187.
$\therefore\ 3\left( \sqrt{3} \right)^{n - 1} = 2187\ \ \Rightarrow 3^{\left( \frac{n}{2} - \frac{1}{2} + 1 \right)} = 3^{7}$
$\Rightarrow \ \frac{n}{2} + \frac{1}{2} = 7 \Rightarrow n = 13$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.