The first two terms of a gp are 4 and 6. What is the 3rd term

Answer 1

In GP:

an = a ∙ r ⁿ ⁻ ¹

where

a = a1 = first term

r = common ratio

a1 = a

a2 = a ∙ r

a2 / a1= a ∙ r / a = r

In this case:

a1 = 4 , a2 = 6

a2 / a1= 6 / 4 = 2 ∙ 3 / 2 ∙ 2 = 3 / 2

r = 3 / 2

a3 = a ∙ r ² = 4 ∙ ( 3 / 2 ) ² = 4 ∙ 9 / 4 = 9

OR

In GP, the common quotient is equal to the quotient of two consecutive terms.

r = a2 / a1 = 6 / 4 = 3 / 2

In GP, each term is the product of the previous term and the common ratio.

a3 = a2 ∙ r = 6 ∙ 3 / 2 = 18 / 2 = 9

Answer 2

r = term(3) / term(2) = 6/4 = 3/2
if term(2) = 6 , then
term(3) = 6(3/2) = 9