Question

interpolar . seis meios aritmeticos entre 14 e 47

Answer 1

Boa tarde Natle!

Solução!

$P.A\{a1-,-,-,-,-,-,an\}$

$a1=14\\\\ n=8\\\\ r=?\\\\ an=47$

Formula da P.A

$an=a1+(n-1).r$

$47=14+(8-1).r\\\\ 47=14+7r\\\\ 33=7r\\\\ r= \dfrac{33}{7}$

Fazendo!


$a1=14\\\\ a2=a1+r=14+ \dfrac{33}{7}= \dfrac{131}{7}\\\\\\ a3=a2+r= \dfrac{131}{7}+ \dfrac{33}{7}= \dfrac{164}{7}\\\\\\\ a4=a3+r= \dfrac{164}{7}+ \dfrac{33}{7}= \dfrac{197}{7}\\\\\\\ a5=a4+r= \dfrac{197}{7}+ \dfrac{33}{7}= \dfrac{230}{7} \\\\\\\ a6=a5+r= \dfrac{230}{7} + \dfrac{33}{7} = \dfrac{263}{7}\\\\\\ a7=a6+r= \dfrac{263}{7} + \dfrac{33}{7}= \dfrac{296}{7}\\\\\\ a8=a7+r= \dfrac{296}{7} + \dfrac{33}{7}= \dfrac{329}{7}=47$


$P.A\{14,\dfrac{131}{7},\dfrac{164}{7},\dfrac{197}{7},\dfrac{230}{7},\dfrac{263}{7},\dfrac{296}{7},47\}$

Boa tarde!
Bons estudos!