Question

These are the first six terms of a sequence with a = 2:

Answer 1

Answer:

$a_{n}$ = 9$a_{n-1}$

Step-by-step explanation:

Note there is a common ratio r between consecutive terms of the sequence, that is

r = 18 ÷ 2 = 162 ÷ 18 = 1458 ÷ 162 = 13122 ÷ 1458 = 9

A recursive formula allows a term in the sequence to be found by multiplying the previous term by the common ratio, that is

$a_{n}$ = 9$a_{n-1}$ with a₁ = 2