Question

pessoal so quero que responda a questao 119. , obg pela atencao .

Answer 1

$Ola'~~~~\mathbf{Diego} \\ \\ ...~;~\underbrace{I}_{a_1}~;~ \underbrace{\frac{ \sqrt{3}-1 }{2}}_{a_2}~;~ \underbrace{\frac{2- \sqrt{3} }{2}}_{a_3}~;.... \\ \\ Se~sabe~que~a~raz\~ao~(q)~e': \\ \\ q= \frac{a_2}{a_1}= \frac{a_3}{a_2} =\ \textgreater \ q= \frac{ \frac{2- \sqrt{3} }{2} }{ \frac{ \sqrt{3}-1 }{2} } ~~--\ \textgreater \ desenvolviendo \\ \boxed{q= \frac{ \sqrt{3}-1 }{2}} \\ \\ ====================================== \\ Se~sabe~que: \\ \boxed{a_n=a_1.q ^{n-1}} \\ \\ a_3= \frac{2- \sqrt{2} }{2}$

$a_1=I=? \\ q= \frac{ \sqrt{3}-1 }{2} \\ \\ Substituindo~en~la~f\acute{o}rmula~tenemos: \\ \frac{2- \sqrt{3} }{2}=I.( \frac{ \sqrt{3} -1}{2}) ^{(3-1)} ~~--\ \textgreater \ desenvolviendo~ \\ \frac{2- \sqrt{3} }{2}=I.[ \frac{3-2 \sqrt{3}+1 }{4}] \\ 4-2 \sqrt{3}=I(4-2 \sqrt{3}) \\ I= \frac{4-2 \sqrt{3} }{4-2 \sqrt{3} } \\\boxed{\boxed{ I=1}} \\ \\ ===================================== \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Espero~ter~ajudado!!$