Question

is 256 a term of the GP {3,6,12,24,....}?​

Answer 1

Answer:

No

Step-by-step explanation:

First term,a = 3

Common ratio,r = 2

nth term of GP, $T_{n} =ar^{n-1}$

                       256=$ar^{n-1}$

                       $3.2^{n-1} =256$

                         $2^{n-1}=\frac{256}{3}$

This cant be solved further.

n should be an integer for it to be a part of GP.