Question

Inserindo oito meios aritméticos entre -2 e 43, termos

Answer 1

Olá,

vamos identificar os termos desta P.A.:

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~10~termos \\P.A.=(\overbrace{-2,\underbrace{\dfrac{a_2}{}, \dfrac{a_3}{~~}, \dfrac{a_4}{~~} , \dfrac{a_5}{~~} ,\dfrac{a_6}{}, \dfrac{a_7}{}, \dfrac{a_8}{}, \dfrac{a_9}{}},43})\\ ~~~~~~~~~~~~~~a_1~~~~~~~~~~~~~~~~8~meios~~~~~~~~~~~~~a_{10}$

Aplicando a fórmula do termo geral da P.A., podemos achar a razão e inserirmos os oito meios aritméticos, somando a razão à partir do 1º termo:

$a_n=a_1+(n-1)r\\ 43=-2+(10-1)\cdot r\\ 43+2=9\cdot r\\ 9r=45\\\\ r= \dfrac{45}{9}\\\\ r=5.\\.$

Então a P.A. com os oito termos inseridos é:

$\large\boxed{\boxed{P.A.=\left(-2, \dfrac{3}{~~~}, \dfrac{8}{~~~} ,\dfrac{13}{}, \dfrac{18}{}, \dfrac{23}{}, \dfrac{28}{}, \dfrac{33}{}, \dfrac{38}{},43\right)}}}.\\.$

Tenha ótimos estudos ;D

Answer 2

Vamos encontrar a razão da PA

an = a1 + ( n -1) * r
43 = -2 + ( 10 -1) * r
43 = -2 + 9 * r
45 = 9r
9r = 45
r = 45/9
r = 5
=================================
Encontrando os termos da PA

n an = a1 + ( n -1) * r     an
1 an = -2 + ( 1 -1)  * 5   =  -2
2 an = -2 + ( 2 -1)  * 5   =   3
3 an = -2 + ( 3 -1) *  5   =   8
4 an = -2 + ( 4 -1)  * 5   =  13
5 an = -2 + ( 5 -1)  * 5   =  18
6 an = -2 + ( 6 -1)  * 5   =  23
7 an = -2 + ( 7 -1)  * 5   =  28
8 an = -2 + ( 8 -1)  * 5   =  33
9 an = -2 + ( 9 -1)  * 5   =  38
10 an = -2 + ( 10 -1)* 5  =  43

PA = (-2, 3, 8, 13, 18, 23, 28, 33, 43)