Question

de a soma dos termos da pg (6,12,...,192)

Answer 1

P.g

Q = a2/a1
Q = 12/6   →  2

$A_{n} = a_{1} . Q^(n-1)$

192 = 6 . 2⁽ⁿ ⁻¹⁾
192/6 = 2⁽ⁿ ⁻¹⁾
32 = 2⁽ⁿ ⁻¹⁾
2⁵ = 2⁽ⁿ ⁻¹⁾

5 + 1 = n
6 = n

$S_{n} = \frac{ a_{1} ( q^{n} -1)}{q -1}$

S6 = 6 . (2⁶ -1)/2 -1
S6 = 6 . (64 -1)
S6 = 6 . 63
S6 = 378