Question

Quantos termos tem a PG (8, 32...2*31)?

Answer 1

razão Q = 32/8 = 4 
AN = A1.Q^(N-1)

Sendo assim:
2^31 = 2³.2²^ (N - 1)
2^31 = 2³.4^ (N-1)
2^31/2³ = 4^ (N-1)
2^28 = 2^2 (N-1)
2(N-1) = 28
2N-2 = 28
2n = 30
n = 15

Answer 2

a₁ = 8; an = 2³¹; r = 4; n = ? 
an= a₁ . qⁿ ⁻ ¹
2³¹ = 8 × 4ⁿ ⁻ ¹
2³¹ = 2³×2²⁽ⁿ ⁻ ¹⁾
2³¹ ÷ 2³ = 2 ⁽²ⁿ ⁻ ²⁾
2²⁸ = 2 ⁽²ⁿ ⁻ ²⁾
28 = 2n - 2⇄ Invertendo a ordem
2n = 28 + 2
n = 30 ÷ 2
n = 15 ← 15 termos.