Question

Qual a soma dos termos da PG: ( 1/3, 1/6,..., 1/96) ?​

Answer 1

Resposta:

S=21/128

Explicação passo-a-passo:

Calculando a razão:

q=a2/a1

q=(1/6)/(1/3)

q=(1/6)(3/1)

q=3/6

q=1/2

an=a1.q^(n-1)

1/96=1/3.[(1/2)^(n-1)]

(1/96)/(1/3)=[(1/2)^(n-1)]

(1/96)/(3/1)=[(1/2)^(n-1)]

1/32=[(2^-1)^(n-1)]

1/2^5=[(2^-1)^(n-1)]

2^-5=[(2^-1)^(n-1)]

5=-(n-1)

-5=-n+1

n=1+5

n=6

S=a1(q^n     -1)/(q-1)

S=1/3[(1/2)^6  -1]/(1/2-1)

S=1/3[[1/64-1]/-1/2

S=1/3[-63/64]/-1/2

S=1/3[63/128]

S=21/128