Question

cincos primeiros termos de pg

a3=9 e q=3
a5= 1 sobre 16 e q= 1 sobre 2
a1=x e q= x sobre 3
a1=m e q=n

Answer 1

PG (__, __, 9, __, __)
a₄ = 9x3=27
a₅ = 27x3=81
a₂ = 9:3=3
a₁ = 3:3 = 1
PG(1; 3; 9; 27; 81)
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PG(__;__;__;__;1/16)
a₄ = 1/16 : 1/2 = 1/16 x 2 = 2/16
a₃ = 2/16 : 1/2 = 2/16 x 2 = 4/16
a₂ = 4/16 : 1/2 = 4/16 x 2 = 8/16
a₁ = 8/16 : 1/2 = 8/16 x 2 = 16/16 = 1
PG(1 ; 8/16;  4/16;  2/16; 1/16)
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a₁ = x
a₂= x.x/3 = x²/3
a₃ = x²/3 . x/3 = x³/9
a₄= x³/9 . x/3 = x⁴/27
a₅= x⁴/27 . x/3 = x⁵/81
PG(x ; x²/3;  x³/9; x⁴/27; x⁵/81)
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a₁=m
a₂=mn
a₃=mn.n = mn²
a₄ = mn².n = mn³
a₅= mn³.n = mn⁴
PG(m; mn; mn²; mn³; mn⁴)