Question

INTERPOLE OU INSIRA doze meios aritméticos entre 5 e 60.

Answer 1

Interpolando 12 meios aritméticos, estaremos formando uma PA com 14 termos sendo que a₁=5 e a₉=60.

            {5 , a₂ , a₃ , a₄ , a₅ , a₆ , a₇ , a₈ , a₉ , a₁₀ , a₁₁ , a₁₂ , a₁₃ , 60}

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

$a_n~=~a_1+(n-1)\cdot r\\\\\\a_{14}~=~a_1~+~(14-1)\cdot r\\\\\\60~=~5~+~13r\\\\\\r~=~\dfrac{60-5}{13}\\\\\\\boxed{r~=~\dfrac{55}{13}}$

Temos então uma PA com 14 termos, razão 55/13, a₁=5 e a₁₄=60.

Podemos ainda determinar os termos que foram interpolados:

$a_2~=~a_1+r~~=~5+\dfrac{55}{13}~~~\rightarrow~~~\boxed{a_2~=~\dfrac{120}{13}}\\\\\\a_3~=~a_2+r~~=~\dfrac{120}{13}+\dfrac{55}{13}~~~\rightarrow~~~\boxed{a_3~=~\dfrac{175}{13}}\\\\\\a_4~=~a_3+r~~=~\dfrac{175}{13}+\dfrac{55}{13}~~~\rightarrow~~~\boxed{a_4~=~\dfrac{230}{13}}\\\\\\a_5~=~a_4+r~~=~\dfrac{230}{13}+\dfrac{55}{13}~~~\rightarrow~~~\boxed{a_5~=~\dfrac{285}{13}}\\\\\\a_6~=~a_5+r~~=~\dfrac{285}{13}+\dfrac{55}{13}~~~\rightarrow~~~\boxed{a_6~=~\dfrac{340}{13}}$

$a_7~=~a_6+r~~=~\dfrac{340}{13}+\dfrac{55}{13}~~~\rightarrow~~~\boxed{a_7~=~\dfrac{395}{13}}\\\\\\a_8~=~a_7+r~~=~\dfrac{395}{13}+\dfrac{55}{13}~~~\rightarrow~~~\boxed{a_8~=~\dfrac{450}{13}}\\\\\\a_9~=~a_8+r~~=~\dfrac{450}{13}+\dfrac{55}{13}~~~\rightarrow~~~\boxed{a_9~=~\dfrac{505}{13}}\\\\\\a_{10}~=~a_9+r~~=~\dfrac{505}{13}+\dfrac{55}{13}~~~\rightarrow~~~\boxed{a_{10}~=~\dfrac{560}{13}}\\\\\\a_{11}~=~a_{10}+r~~=~\dfrac{560}{13}+\dfrac{55}{13}~~~\rightarrow~~~\boxed{a_{11}~=~\dfrac{615}{13}}$

$a_{12}~=~a_{11}+r~~=~\dfrac{615}{13}+\dfrac{55}{13}~~~\rightarrow~~~\boxed{a_{12}~=~\dfrac{670}{13}}\\\\\\a_{13}~=~a_{12}+r~~=~\dfrac{670}{13}+\dfrac{55}{13}~~~\rightarrow~~~\boxed{a_{13}~=~\dfrac{725}{13}}$