Question

Na PG (1/2,-1,2,-4,.....), encontre os termos a9 e a10.
A) 32 e-64
B) 64 e-128
C) -64 e 128
D) 128 e-256
E) 256 e -512​

me ajudem pfvrrr

Answer 1

Resposta:

$D)\ 128\ e\ -256$

Explicação passo a passo:

$a_{1}=\frac{1}{2} \\\\a_{3} =2 \\\\a_{4} =-4\\\\q=\frac{a_{4}}{a_{3}} =\frac{-4}{2} =-\frac{4}{2} =-2$

$a_{n} =a_{1}\times q^{n-1}$

$Para$ $n=9:$

$a_{9} =\frac{1}{2} \times (-2)^{9-1}\\\\a_{9} =\frac{1}{2} \times (-2)^{8}\\\\a_{9} =\frac{1}{2} \times 256\\\\a_{9} =\frac{1 \times 256}{2}\\\\a_{9} =\frac{256}{2}\\\\a_{9} =128\\\\\\\\a_{n} =a_{n-1} \times q\\\\\\Para \ n=10:\\\\\\a_{10} =a_{10-1} \times (-2)\\\\a_{10} =a_{9} \times (-2)\\\\a_{10} =128 \times (-2)\\\\a_{10} =-256$