Question

o numero de termos da pg (1,3,9...) sedo an= 729 é:

Answer 1

PG
a1 = 1
an = 729
q = 3 : 1 = 3
n =?
An = a1 . $q^{n - 1}$
729 = 1 . $3^{n - 1}$
729 = $3^{n - 1}$
$3^{6}$ = $3^{n - 1}$
cancela a base 3
6 = n - 1
6 + 1 = n
n = 7

PG = ( 1, 3, 9, 27, 81, 243, 729)

Answer 2

An = 729

a₁ = 1

q = a₂ / a₁

q = 3/1

q = 3

n = ?

An = a₁ * qⁿ⁻¹

729 = 1 * 3ⁿ⁻¹

3ⁿ⁻¹ = 729 / 1

3ⁿ⁻¹ = 729

3ⁿ⁻¹ = 3⁶

n - 1 = 6

n = 1 + 6  

n = 7