Question

Me ajudem por favor
calcule o n° de termos da P.G ( 1/1024, 1/512, 1/256, .. 2 elevado a 31 )

Answer 1

$a_{1}=\dfrac{1}{1024}=\dfrac{1}{2^{10}}=2^{-10}\\\\\\a_{2}=\dfrac{1}{512}=\dfrac{1}{2^{9}}=2^{-9}$

Achando a razão da P.G:

$q=\dfrac{a_{2}}{a_{1}}=\dfrac{2^{-9}}{2^{-10}}=2^{-9-(-10)}=2$
_________________________

$a_{n}=a_{1}\cdot q^{n-1}\\2^{31}=2^{-10}\cdot2^{n-1}\\2^{31+10}=2^{n-1}\\2^{n-1}=2^{41}\\n-1=41\\\\\boxed{\boxed{n=42}}$