Question

determine o número de termos da progressão geométrica:

Answer 1

$\large O ~n\acute{u}mero ~de ~termos ~da ~PG ~ \Rightarrow ~n = 7$

Encontrar a razão da PG

$q = \dfrac{a2}{a1}\\\\\\q = \dfrac{5}{25}\\\\\\q = \dfrac{1}{5}$

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Encontrar o número de termos da PG

$an = a1 . (\dfrac{1}{5}) ^{n - 1}\\\\\\ \dfrac{1}{625} =25 ~. ~\dfrac{1}{5}^{n - 1}\\\\\\\dfrac{ \dfrac{1}{625}}{25} =\dfrac{1}{5}^{n - 1}\\\\\\\dfrac{1}{15625} =\dfrac{1}{5}^{n - 1}\\\\\\\ \dfrac{1}{5^6} =\dfrac{1}{5}^{n - 1}\\\\\\6 = n - 1\\\\\\n = 6 + 1\\\\\\n = 7$

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Answer 2

Resolução!

■ Progressão Geométrica

q = a2/a1

q = 5/25

q = 1/5

an = a1 * q^n - 1

1/625 = 25 * (1/5)^n - 1

1/615 / 25 = (1/5)^n - 1

1/625 * 1/25 = (1/5)^n - 1

1/15625 = (1/5)^n - 1

(1/5)^6 = (1/5)^n - 1

n - 1 = 6

n = 6 + 1

n = 7

Resposta: PG de 7 termos

Espero ter ajudado