Question

Interpolar 5 meios geométricos entre -1/4 e -16

Answer 1

Interpolando 5 meios geométricos, estaremos formando uma PG com 7 termos sendo que a₁=-1/4 e a₇=-16.

                                 {-1/4 , a₂ , a₃ , a₄ , a₅ , a₆ , -16}

Vamos então começar determinando a razão "q" dessa PG com auxilio da equação do termo geral da PG:

$a_n~=~a_1\cdot q^{n-1}\\\\\\a_7~=~a_1\cdot q^{7-1}\\\\\\-16~=\,-\dfrac{1}{4}\cdot q^{6}\\\\\\q^6~=~\dfrac{-16}{-\frac{1}{4}}\\\\\\q^{6}~=~16\cdot\dfrac{4}{1}\\\\\\q^{6}~=~64\\\\\\q^{6}~=~2^6\\\\\\q^{6\!\!\!\backslash}~=~2^{6\!\!\!\backslash}\\\\\\\boxed{q~=~2}$

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

$a_2~=~a_1\cdot q^{2-1}~=\,-\dfrac{1}{4}\cdot 2^1~~~~\longrightarrow~~~~\boxed{a_2~=\,-\dfrac{1}{2}}\\\\\\\\a_3~=~a_2\cdot q^{3-2}~=\,-\dfrac{1}{2}\cdot 2^1~~~~\longrightarrow~~~~\boxed{a_3~=\,-1}\\\\\\\\a_4~=~a_3\cdot q^{4-3}~=\,-1\cdot 2^1~~~~\longrightarrow~~~~\boxed{a_4~=\,-2}\\\\\\\\a_5~=~a_4\cdot q^{5-4}~=\,-2\cdot 2^1~~~~\longrightarrow~~~~\boxed{a_5~=\,-4}\\\\\\\\a_6~=~a_5\cdot q^{6-5}~=\,-4\cdot 2^1~~~~\longrightarrow~~~~\boxed{a_6~=\,-8}$

A PG fica: {-1/4 , -1/2 , -1 , -2 , -4 , -8 , -16}