Question

Calcule x, sendo: 5x + x/2 + x/4 + x/8 +.... = 60


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Answer 1

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$\Large\boxed{\sf{\underline{Soma~dos~termos~da~PG~infinita}}}\\\huge\boxed{\boxed{\boxed{\boxed{\sf S_n=\dfrac{a_1}{1-q}}}}}$

$\sf 5x+\dfrac{x}{2}+\dfrac{x}{4}+\dfrac{x}{8}+...=60\\\sf 5x+x\cdot(\underbrace{\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...}_{soma~dos~termos~da~PG~infinita})=60\\\sf a_1=\frac{1}{2}~~a_2=\frac{1}{4}\\\sf q=\dfrac{a_2}{a_1}\\\sf q=\dfrac{\frac{1}{4}}{\frac{1}{2}}\\\sf q=\dfrac{1}{\diagup\!\!\!4_2}\cdot\diagup\!\!\!2=\dfrac{1}{2}$

$\sf S_n=\dfrac{a_1}{1-q}\\\sf S_n=\dfrac{\frac{1}{2}}{1-\frac{1}{2}}\\\sf S_n=\dfrac{\frac{1}{2}}{\frac{2-1}{2}}\\\sf S_n=\dfrac{\frac{1}{2}}{\frac{1}{2}}=1$

$\sf retornando~a~equac_{\!\!,}\tilde{a}o~temos:\\\sf 5x+\dfrac{x}{2}+\dfrac{x}{4}+\dfrac{x}{8}+...=60\\\sf 5x+x\cdot1=60\\\sf 6x=60\\\sf x=\dfrac{60}{6}\\\huge\boxed{\boxed{\boxed{\boxed{\sf x=10\checkmark}}}}$