Question

Em relação à sequência numérica (4, 8, 12, 16, 20, …), determine:

Answer 1

Explicação passo-a-passo:

a)

$\sf r=a_2-a_1$

$\sf r=8-4$

$\sf \red{r=4}$

b)

$\sf a_n=a_1+(n-1)\cdot r$

$\sf a_{15}=a_1+14r$

$\sf a_{15}=4+14\cdot4$

$\sf a_{15}=4+56$

$\sf \red{a_{15}=60}$

c)

• $\sf a_n=a_1+(n-1)\cdot r$

$\sf a_{10}=a_1+9r$

$\sf a_{10}=4+9\cdot4$

$\sf a_{10}=4+36$

$\sf a_{10}=40$

• $\sf a_n=a_1+(n-1)\cdot r$

$\sf a_{20}=a_1+19r$

$\sf a_{20}=4+19\cdot4$

$\sf a_{20}=4+76$

$\sf a_{20}=80$

Logo:

$\sf a_{10}+a_{20}=40+80$

$\sf \red{a_{10}+a_{20}=120}$

d)

$\sf a_n=a_1+(n-1)\cdot r$

$\sf a_{152}=a_1+151r$

$\sf a_{152}=4+151\cdot4$

$\sf a_{152}=4+604$

$\sf a_{152}=608$

• $\sf a_n=a_1+(n-1)\cdot r$

$\sf a_{88}=a_1+87r$

$\sf a_{88}=4+87\cdot4$

$\sf a_{88}=4+348$

$\sf a_{88}=352$

Logo:

$\sf \dfrac{a_{152}+a_{88}}{3}=\dfrac{608+352}{3}$

$\sf \dfrac{a_{152}+a_{88}}{3}=\dfrac{960}{3}$

$\sf \red{\dfrac{a_{152}+a_{88}}{3}=320}$