Question

Qual o oitavo termo da P.A (1, 3/2, 2, ...)?

Answer 1

$a_{1} =1$

$r= \frac{3}{2} -1= \frac{3}{2} - \frac{2}{2} = \frac{1}{2}$

$a_{8} =a_{1} +7.r=1+7. \frac{1}{2} = \frac{2}{2} + \frac{7}{2} = \frac{9}{2}$

Answer 2

Temos o  primeiro termo

$a_1=1$

Determinando a razão

$r=a_2-a_1$

$r= \frac{3}{2} - \frac{2}{2} ~\to~r= \frac{1}{2}$

8° termo

$an=a_1+(n-1)r\\a_{8}=1+(8-1)* \frac{1}{2} \\a_8=1+ \frac{7}{2} \\a_8= \frac{9}{2}$