Question

A razão entre a soma dos termos da PA(4,10,16,. , 940) e a soma dos termos da PG(2, 4, 8,. , 4096) é aproximadamente:


a. 6

b. 8

c. 7

d. 9

e. 10.

Answer 1

Resposta:

d. 9

Explicação passo a passo:

$a_{1} =4$

$a_{2} =10$

$r=a_{2} -a_{1} =10-4=6$

$a_{n} =940$

$n=?$

$S_{n} =?$

$a_{1} +(n-1).r=a_{n}$

$(n-1).r=a_{n}-a_{1}$

$n-1=\frac{a_{n}-a_{1}}{r}$

$n=\frac{a_{n}-a_{1}}{r}+1$

$n=\frac{a_{n}-a_{1}}{r}+1.\frac{r}{r}$

$n=\frac{a_{n}-a_{1}}{r}+\frac{r}{r}$

$n=\frac{a_{n}-a_{1}+r}{r}$

$n=\frac{940-4+6}{6}$

$n=\frac{942}{6}$

$n=157$

$S_{n} =\frac{(a_{1} +a_{n} ).n}{2}$

$S_{157} =\frac{(4 +940 ).157}{2}$

$S_{157} =\frac{944 .157}{2}$

$S_{157} =\frac{148208}{2}$

$S_{157} =74104$

$a_{1} =2$

$a_{2} =4$

$q=\frac{a_{2}}{a_{1}} =\frac{4}{2} =2$

$a_{n} =4096$

$n=?$

$S_{n} =?$

$a_{1} .q^{n-1}=a_{n}$

$q^{n-1}=\frac{a_{n}}{a_{1}}$

$q^{n}.q^{-1}=\frac{a_{n}}{a_{1}}$

$q^{n}.\frac{1}{q}=\frac{a_{n}}{a_{1}}$

$q^{n}=\frac{\frac{a_{n}}{a_{1}}}{\frac{1}{q} }$

$q^{n}=\frac{a_{n} .q}{a_{1} }$

$2^{n}=\frac{4096 .2}{2 }$

$2^{n}=4096$

$2^{n}=2^{12}$

$n=12$

$S_{n} =\frac{a_{1}.(q^{n}-1) }{q-1}$

$S_{12} =\frac{2.(2^{12}-1) }{2-1}$

$S_{12} =\frac{2.(4096-1) }{1}$

$S_{12} =2.4095$

$S_{12} =8190$

$\frac{S_{157} }{S_{12} } =\frac{74104}{8190}$

≈ $9$