Question

Quantos termos tem a PG (2, 4, 8, ..., 1024)?

Answer 1

Primeiro vamos encontrar a razão dessa P.G :

$\mathtt{q = A_2 \div A_1} \\ \mathtt{q = 4 \div 2} \\ \mathtt{q = 2}$

Agora vamos aplicar o termo geral dessa equação :

$\mathtt{A_n = A_1 ~.~q^{n-1}} \\ \mathtt{1.024 = 2~.~2^{n-1}} \\ \mathtt{2^{n-1} = 1.024\div2} \\ \mathtt{2^{n-1} =512 }$

vamos fator 512

512 |2

256 |2

128 |2

64 |2

32 |2

16 |2

8 |2

4 |2

2 |2

1

2 . 2 . 2 . 2 . 2 . 2 . 2 . 2 . 2

2⁹

$\mathtt{2^{n-1} =512 } \\ \mathtt{2^{n-1} = 2^9} \\ \\ \mathtt{corta ~base~igual} \\ \\ \mathtt{n - 1 = 9} \\ \mathtt{n = 9 + 1} \\ \mathtt{n = 10}$

Resposta: 10 termos