Question

Calcule a soma dos infinitos termos de cada PG a seguir.

a) (12, 6, 3, ...)

d) (16, - 4,1,...)

b) (1, 3, 25, ...)

e) (1000, 20,...)

c) (-24;-2,4;-0,24;...)​

Answer 1

Resposta:

a) S = 24

b) S = 5/4

c) S = -240/9

d) S = 64/5

e) S = 1250

Explicação passo a passo:

A razão (q) da PG

q = a₂/a₁

A soma de infinitos termos: S = a₁/(1-q)

a) (12, 6, 3, ...)

a₁ = 12

q = 6/12 = 6÷6/12÷6 = 1/2

S = 12/(1-1/2) = 12/(1/2) = 12.(2/1) = 24

b) (1, 1/5, 1/25, ...)

a₁ = 1

q = (1/5)/1 = 1/5

S = 1/(1-1/5) = 1/[(5-1)/5] = 1/(4/5) = 1.5/4 = 5/4

c) (-24, -2,4, -0,24,...)​

a₁ = -24

q = -2,4/(-24) = 0,1

S = -24/(1-0,1) = -24/0,9 = -24/(9/10) =  -24.(10/9) = -240/9

d) (16, - 4,1,...)

a₁ = 16

q = -4/16 = -4÷4/16÷4 = -1/4

S = 16/[1-(-1/4)] = 16/[(4+1)/4] = 16/[5/4] = 16.(4/5) = 64/5

e) (1000, 20,...)

a₁ = 1000

q = 20/100 =2÷2/10÷2 =1/5

S = 1000/(1-1/5) = 1000/[(5-1)/5] = 1000/(4/5) = 1000.(5/4) = 1250