1. (Unitau) O valor da soma
“S=4+\left (\frac{1}{10} \right )+\left (\frac{36}{10^3}+\frac{36}{10^5}+\frac{36}{10^7}+\frac{36}{10^9}+… \right )”)](https://www.codecogs.com/eqnedit.php?latex=S=4+\left&space;(\frac{1}{10}&space;\right&space;)+\left&space;(\frac{36}{10^3}+\frac{36}{10^5}+\frac{36}{10^7}+\frac{36}{10^9}+…&space;\right&space;)) é igual a:
a) 99/22.
b) 91/22.
c) 91/21.
d) 90/21.
e) 81/23.
2. (Uece) Seja (b1, b2, b3, b4) uma progressão geométrica de razão 1/3. Se b1 + b2 + b3 + b4 = 20, então b4 é igual a:
a) 1/2
b) 3/2
c) 5/2
d) 7/2
1. B
Solução Passo-a-Passo:
“S=4+\left (\frac{1}{10} \right )+\left (\frac{36}{10^3}+\frac{36}{10^5}+\frac{36}{10^7}+\frac{36}{10^9}+… \right )”)](https://www.codecogs.com/eqnedit.php?latex=S=4+\left&space;(\frac{1}{10}&space;\right&space;)+\left&space;(\frac{36}{10^3}+\frac{36}{10^5}+\frac{36}{10^7}+\frac{36}{10^9}+…&space;\right&space;))
“S=4+\left (\frac{1}{10} \right )+36.\left (\frac{1}{10^3}+\frac{1}{10^5}+\frac{1}{10^7}+\frac{1}{10^9}+… \right )”)](https://www.codecogs.com/eqnedit.php?latex=S=4+\left&space;(\frac{1}{10}&space;\right&space;)+36.\left&space;(\frac{1}{10^3}+\frac{1}{10^5}+\frac{1}{10^7}+\frac{1}{10^9}+…&space;\right&space;))
\Rightarrow pg , infinita,, de , q=10^{-2} , e , a1=\frac{1}{10^3}.”)](https://www.codecogs.com/eqnedit.php?latex=S’=\left&space;(\frac{1}{10^3}+\frac{1}{10^5}+\frac{1}{10^7}+…&space;\right&space;)&space;\Rightarrow&space;pg&space;\,&space;infinita,\,&space;de&space;\,&space;q&space;=&space;10^{-2}&space;\,&space;e&space;\,&space;a1=\frac{1}{10^3}.)
\Rightarrow S’=\frac{1}{990}”)](https://www.codecogs.com/eqnedit.php?latex=S’=\frac{a_1}{q-1}\Rightarrow&space;S’=\frac{\frac{1}{10^3}}{1-\frac{1}{10^2}}&space;=&space;\frac{\frac{1}{10^3}}{\frac{99}{10^2}}\Rightarrow&space;\frac{1}{10^3}.\left&space;(&space;\frac{10^2}{99}&space;\right&space;)\Rightarrow&space;S’=\frac{1}{990})
+36.\left&space;(\frac{1}{990}&space;\right&space;)=4+\left&space;(\frac{1}{10}&space;\right&space;)&plus\frac{4}{110} “S=4+\left (\frac{1}{10} \right )+36.S’\Rightarrow S=4+\left (\frac{1}{10} \right )+36.\left (\frac{1}{990} \right )=4+\left (\frac{1}{10} \right )+\frac{4}{110}”)](https://www.codecogs.com/eqnedit.php?latex=S=4+\left&space;(\frac{1}{10}&space;\right&space;)+36.S’\Rightarrow&space;S=4+\left&space;(\frac{1}{10}&space;\right&space;)+36.\left&space;(\frac{1}{990}&space;\right&space;)=4+\left&space;(\frac{1}{10}&space;\right&space;)+\frac{4}{110})
2. A
Solução Passo-a-Passo:
A soma dos quatro primeiros termos é 20 e q = 1/3. Assim:
^4-1&space;\right&space;)}{\left&space;(&space;\frac{1}{3}&space;\right&space;)-1}\Rightarrow&space;20&space;=&space;\frac{b1.\left&space;(&space;\frac{1}{81}&space;\right&space;-1)}{\frac{1-3}{3}} “Sn=\frac{a1.(q^n-1)}{q-1}\Rightarrow S4=\frac{b1.\left (\left ( \frac{1}{3} \right )^4-1 \right )}{\left ( \frac{1}{3} \right )-1}\Rightarrow 20=\frac{b1.\left ( \frac{1}{81} \right -1)}{\frac{1-3}{3}}”)](https://www.codecogs.com/eqnedit.php?latex=S_n=\frac{a_1.(q^n-1)}{q-1}\Rightarrow&space;S_4=\frac{b_1.\left&space;(\left&space;(&space;\frac{1}{3}&space;\right&space;)^4-1&space;\right&space;)}{\left&space;(&space;\frac{1}{3}&space;\right&space;)-1}\Rightarrow&space;20&space;=&space;\frac{b_1.\left&space;(&space;\frac{1}{81}&space;\right&space;-1)}{\frac{1-3}{3}})
}=\frac{40b1}{27}\Rightarrow&space;1&space;=&space;\frac{2b1}{27}\Rightarrow&space;b1=\frac{27}{2} “20=\frac{b1.\left ( \frac{1-81}{81} \right )}{\frac{-2}{3}}\Rightarrow 20=-\frac{80b1}{81}.\frac{3}{\left (-2 \right )}=\frac{40b1}{27}\Rightarrow 1=\frac{2b1}{27}\Rightarrow b1=\frac{27}{2}”)](https://www.codecogs.com/eqnedit.php?latex=20&space;=&space;\frac{b_1.\left&space;(&space;\frac{1-81}{81}&space;\right&space;)}{\frac{-2}{3}}\Rightarrow&space;20=-\frac{80b_1}{81}.\frac{3}{\left&space;(-2&space;\right&space;)}=\frac{40b_1}{27}\Rightarrow&space;1&space;=&space;\frac{2b_1}{27}\Rightarrow&space;b_1=\frac{27}{2})
^3\Rightarrow b4=\frac{27}{2}.\frac{1}{27}\Rightarrow b4=\frac{1}{2}.”)](https://www.codecogs.com/eqnedit.php?latex=b_n=b_1.q^{n-1}\Rightarrow&space;b_4=b_1.q^{4-1}=\frac{27}{2}.\left&space;(\frac{1}{3}&space;\right&space;)^3\Rightarrow&space;b_4=\frac{27}{2}.\frac{1}{27}\Rightarrow&space;b_4&space;=\frac{1}{2}.)
