Question

numa P.A tem-se a1=-3 e r=3/2,calcule: a)a16 b)a39

Answer 1

 

 

a1=-3

r=3/2

an = a1 + (n -1).r

 

a) a16 = -  3 + (16 - 1)(3/2)

             = - 3 + 15x3/2

             = - 3 + 45 / 2

             = - 6/2 + 45/2

     a16 = 39/2

 

b) a39 = -3 + (39 - 1)(3/2

             = -3 + 38x3/2

             = -3 + 57

   a39 = 54

   

Answer 2

Simples, basta jogar na fórmula do termo geral da P.A., que se dá: $<var>\boxed{a_{n} = a_{1} + (n-1) \cdot r}</var>$

 

a) $<var>a_{n} = a_{1} + (n-1) \cdot r</var>$

 

$<var>a_{16} = -3 + (16-1) \cdot \frac{3}{2}</var>$

 

$<var>a_{16} = -3 + 15 \cdot \frac{3}{2}</var>$

 

$<var>a_{16} = -3 + \frac{15 \cdot3}{2}</var>$

 

$<var>a_{16} = -3 + \frac{45}{2}</var>$

 

$<var>a_{16} = -\frac{6}{2} + \frac{45}{2}</var>$

 

$<var>\boxed{a_{16} = \frac{39}{2}} \ ou \ \boxed{19,5}</var>$

 

 

b) $<var>a_{n} = a_{1} + (n-1) \cdot r</var>$

 

$<var>a_{39} = -3 + (39-1) \cdot \frac{3}{2}</var>$

 

$<var>a_{39} = -3 + 38 \cdot \frac{3}{2}</var>$

 

$<var>a_{39} = -3 + \frac{38 \cdot 3}{2}</var>$

 

$<var>a_{39} = -3 + \frac{114}{2}</var>$

 

$<var>a_{39} = -\frac{6}{2} + \frac{114}{2}</var>$

 

$<var>a_{39} = \frac{108}{2}</var>$

 

$<var>\boxed{a_{39} = 54}</var>$