Question

en una progresion geometrica el primer termino es 3/4 el ultimo es 2/9 y la suma de los terminos es de 65/36 ¿de cuantos terminos de compone la progresion ?

Answer 1

a1 = 3/4

an = 2/9

Sn = 65/36

$an = a1*r^{n-1}$

$Sn = \frac{(an*r-a1)}{r-1}$

65/36 = [(2/9)r - (3/4)](r - 1)

65(r - 1) = 36[(2/9)r- (3/4)]

65r - 65 = 8r - 27

65r - 8r = - 27 + 65

57r = 38

r = 38/57

r = 2/3.

Ahora en $an = a1*r^{n-1}$

Donde n = Numero de terminos

$2/9 = (3/4)*(2/3)^{n-1}$

$\frac{2/9}{3/4}=(2/3)^{n-1}$

$\frac{8}{27}=(2/3)r^{n-1}$

Aplico logaritmo natural Ln en ambos lados

Ln(8/27) = Ln[(2/3)^(n - 1)]

Ln(8/27) = (n-1)Ln(2/3)

n - 1 = [Ln(8/27)]/[Ln2/3)]

n - 1 = 3

n = 3 + 1

n = 4.

El numero de terminos es 4.

Probemos para n = 4, r = 2/3, a1 = 3/4

a4 = (3/4)(2/3)^(4 - 1)

a4 = (3/4)[(2/3)^(3)]

a4 = (3/4)[8/27]

a4 = 2/9

Rta: La progresion tiene 4 terminos