Question

Determine o numero de termos da PG (3125,625.....,1/25)

Answer 1

a1 = 3125

a2 = 625

q = 625/3125 = 1/5 ****

an = 1/25

an = a1.qⁿ⁻¹

125 = 3125 . ( 1/5)ⁿ⁻¹

( 1/5)ⁿ⁻¹ = 125/3125 = 1/25 = ( 1/5²) ou ( 1/5)² ***

( 1/5)ⁿ⁻¹ = ( 1/5)²

n - 1 = 2

n = 2 + 1 = 3 ***** resposta