Question

Insira 12 meios aritméticos entre -3/4 e 11/6.​

Answer 1

Interpolando 7 meios aritméticos, estaremos formando uma PA com 14 termos sendo que a₁=-3/4 e a₁₄=11/6.

              {-3/4 , a₂ , a₃ , a₄ , a₅ , a₆ , a₇ , a₈ , a₉ , a₁₀ , a₁₁ , a₁₂ , a₁₃ , 11/6}

Vamos então começar determinando a razão r dessa PA com auxilio da equação do termo geral da PA:

$\boxed{a_n~=~a_1+(n-1)\cdot r}\\\\\\a_{14}~=~a_1+(14-1)\cdot r\\\\\\\dfrac{11}{6}~=\,-\dfrac{3}{4}+13r\\\\\\13r~=~\dfrac{11}{6}+\dfrac{3}{4}\\\\\\13r~=~\dfrac{2\cdot11+3\cdot3}{12}\\\\\\13r~=~\dfrac{22+9}{12}\\\\\\13r~=~\dfrac{31}{12}\\\\\\\boxed{r~=~\dfrac{31}{156}}$

Podemos agora determinar os termos que foram interpolados utilizando, novamente, a equação do termo geral:

$a_2~=~a_1+r~=\,-\dfrac{3}{4}+\dfrac{31}{156}~~~\rightarrow~~~\boxed{a_2~=\,-\dfrac{43}{78}}\\\\\\a_3~=~a_2+r~=\,-\dfrac{43}{78}+\dfrac{31}{156}~~~\rightarrow~~~\boxed{a_3~=\,-\dfrac{55}{156}}\\\\\\a_4~=~a_3+r~=\,-\dfrac{55}{156}+\dfrac{31}{156}~~~\rightarrow~~~\boxed{a_4~=\,-\dfrac{2}{13}}\\\\\\a_5~=~a_4+r~=\,-\dfrac{2}{13}+\dfrac{31}{156}~~~\rightarrow~~~\boxed{a_5~=~\dfrac{7}{156}}\\\\\\a_6~=~a_5+r~=~\dfrac{7}{156}+\dfrac{31}{156}~~~\rightarrow~~~\boxed{a_6~=~\dfrac{19}{78}}$

$a_7~=~a_6+r~=~\dfrac{19}{78}+\dfrac{31}{156}~~~\rightarrow~~~\boxed{a_7~=~\dfrac{23}{52}}\\\\\\a_8~=~a_7+r~=~\dfrac{23}{52}+\dfrac{31}{156}~~~\rightarrow~~~\boxed{a_8~=~\dfrac{25}{39}}\\\\\\a_9~=~a_8+r~=~\dfrac{25}{39}+\dfrac{31}{156}~~~\rightarrow~~~\boxed{a_9~=~\dfrac{131}{156}}\\\\\\a_{10}~=~a_9+r~=~\dfrac{131}{156}+\dfrac{31}{156}~~~\rightarrow~~~\boxed{a_{10}~=~\dfrac{27}{26}}\\\\\\a_{11}~=~a_{10}+r~=~\dfrac{27}{26}+\dfrac{31}{156}~~~\rightarrow~~~\boxed{a_{11}~=~\dfrac{193}{156}}$

$a_{12}~=~a_{11}+r~=~\dfrac{193}{156}+\dfrac{31}{156}~~~\rightarrow~~~\boxed{a_{12}~=~\dfrac{56}{39}}\\\\\\a_{13}~=~a_{12}+r~=~\dfrac{56}{39}+\dfrac{31}{156}~~~\rightarrow~~~\boxed{a_{13}~=~\dfrac{85}{52}}$