Geometric progression

Question

The first and three more successive terms of a geometric progression are given as follows:6...162,y,1458..where y>0

(a) Obtain the common ratio and the tenth term

(b) find the smallest integers, n such that the n th term exceeds 20 000

Answer 1

y is the geometric mean between 162 and 1458

y² = (162)(1458)

y = 486

r = 486 / 162

r = 3

a₁₀ = 6(3)⁹

20000 < 6(3)^n-1