Question

qual o valor de x na equacao 2x + x/5 +x/5 + ... = 20/3 dado que (2x, x/5 , x/50...sao termos de uma progressao geometrica infinita? a 4 b 2 c 3 d 5 e 6

Answer 1

Se $2x,\,\frac{x}{5},\,\frac{x}{50},\,\dots$ são termos de uma P.G infinita, calcularemos a sua razão.

Claramente $x\neq0$, pois $0\neq\frac{20}{3}$, então

$q=\dfrac{a_{2}}{a_{1}}=\dfrac{(\frac{x}{5})}{2x}=\dfrac{x}{5\cdot2x}=\dfrac{1}{10}$

Como $|q|=|\frac{1}{10}|=\frac{1}{10}\,\,\textless\,\,1$, a soma infinita dos termos dessa P.G é convergente, e dada por

$S_{\infty}=a_{1}+a_{2}+a_{3}+\dots=\dfrac{a_{1}}{1-q}$

Como $a_{1}=2x,\,q=\frac{1}{10}$, temos que

$2x+\dfrac{x}{5}+\dfrac{x}{50}+\dots=\dfrac{a_{1}}{1-q}\\\\\\2x+\dfrac{x}{5}+\dfrac{x}{50}+\dots=\dfrac{2x}{1-\frac{1}{10}}\\\\\\2x+\dfrac{x}{5}+\dfrac{x}{50}+\dots=\dfrac{2x}{(\frac{9}{10})}\\\\\\2x+\dfrac{x}{5}+\dfrac{x}{50}+\dots=2x\cdot\dfrac{10}{9}\\\\\\\boxed{\boxed{2x+\dfrac{x}{5}+\dfrac{x}{50}+\dots=\dfrac{20}{9}\cdot x}}$

Agora podemos encontrar o valor de x:

$2x+\dfrac{x}{5}+\dfrac{x}{50}+\dots=\dfrac{20}{3}\\\\\\\dfrac{20}{9}\cdot x=\dfrac{20}{3}\\\\\\x=\dfrac{20}{3}\cdot\dfrac{9}{20}\\\\\\x=\dfrac{9}{3}\\\\\\\boxed{\boxed{x=3}}$