Question

Interpale 11 meios aritméticos entre 16 e 20 e calcule o valor de a+a+a

Answer 1

Interpolando 11 termos entre os 2 já existentes ficaremos com 13 no total.

A PA fica:

16 , a2 , a3 , a4 , a5 , a6 , a7 , a8 , a9 , a10 , a11 , a12 , 20

Utilizando a equação do termo geral, podemos achar o valor da razão:

$a_n~=~a_1+(n-1).r\\\\\\a_{13}~=~a_1+(13-1).r\\\\\\20~=~16+12r\\\\\\12r~=~20-16\\\\\\r~=~\frac{4}{12}\\\\\\\boxed{r~=~\frac{1}{3}}$

Podemos agora achar o valor dos termos adicionados:

$a_2~=~a_1+r~=~16+\frac{1}{3}~~\rightarrow~~\boxed{a_2~=~\frac{49}{3}}\\\\\\a_3~=~a_2+r~=~\frac{49}{3}+\frac{1}{3}~~\rightarrow~~\boxed{a_3~=~\frac{50}{3}}\\\\\\a_4~=~a_3+r~=~\frac{50}{3}+\frac{1}{3}~=~\frac{51}{3}~~\rightarrow~~\boxed{a_4~=~17}\\\\\\a_5~=~a_4+r~=~\frac{51}{3}+\frac{1}{3}~~\rightarrow~~\boxed{a_5~=~\frac{52}{3}}\\\\\\a_6~=~a_5+r~=~\frac{52}{3}+\frac{1}{3}~~\rightarrow~~\boxed{a_6~=~\frac{53}{3}}$

$a_7~=~a_6+r~=~\frac{53}{3}+\frac{1}{3}~=~\frac{54}{3}~~\rightarrow~~\boxed{a_7~=~18}\\\\\\a_8~=~a_7+r~=~\frac{54}{3}+\frac{1}{3}~~\rightarrow~~\boxed{a_8~=~\frac{55}{3}}\\\\\\a_9~=~a_8+r~=~\frac{55}{3}+\frac{1}{3}~~\rightarrow~~\boxed{a_9~=~\frac{56}{3}}\\\\\\a_{10}~=~a_9+r~=~\frac{56}{3}+\frac{1}{3}~=~\frac{57}{3}~~\rightarrow~~\boxed{a_{10}~=~19}\\\\\\a_{11}~=~a_{10}+r~=~\frac{57}{3}+\frac{1}{3}~~\rightarrow~~\boxed{a_{11}~=~\frac{58}{3}}$

$a_{12}~=~a_{11}+r~=~\frac{58}{3}+\frac{1}{3}~~\rightarrow~~\boxed{a_{12}~=~\frac{59}{3}}$