Question

what is the common ratio of a geometric progression were the first term is 3 and the 4th term is 648​

Answer 1

Answer:

$r = 6$

Step-by-step explanation:

Given

$a = 3$ --- First, term

$T_4 = 648$ --- Fourth term

Required: Determine the common ratio

The nth term of a GP is

$T_n = ar^{n-1$

Substitute 4 for n

$T_4 = ar^{4-1$

$T_4 = ar^{3$

Substitute 648 and 3 for T4 and 3 respectively.

$648 = 3 * r^{3$

$r^3 = 216$

Take cube roots of both sides

$r =\sqrt[3]{216}$

$r = 6$

The common ratio is 6