Question

Dada a p.g (2 4 8,....4096),quantos termos tem a p.g o valor de a8

Answer 1

$\text{PG}(2,4,8,\dots,4096)$

A razão é $q=\dfrac{4}{2}~\longrightarrow~q=2$

$\bullet~~\text{O valor de}~a_8$

$a_n=a_1\cdot q^{n-1}$

$a_8=2\cdot2^{7}$

$a_8=2\cdot128$

$\boxed{a_8=256}$

$\bullet~~\text{N}\acute{u}\text{mero}~\text{de}~\text{termos}$

$a_n=a_1\cdot q^{n-1}$

$4096=2\cdot2^{n-1}$

$\dfrac{4096}{2}=2^{n-1}$

$2048=2^{n-1}$

$2^{11}=2^{n-1}$

Igualando os expoentes:

$11=n-1~\longrightarrow~n=11+1~\longrightarrow~\boxed{n=12}$