Question

em uma P.A A3=17 e a20=119 qual o 12​

Answer 1

Resposta:

$razao=a_{2}-a_{1}\\razao=8-3\\razao=5\\\\a_{16}=a_{1}+(n-1)*r\\a_{16}=3+(16-1)*5\\a_{16}=3+(15*5)\\a_{16}=3+75\\\boxed {a_{16}=78} \\\\\\a_{3}=17\\a_{20}=119\\a_{12}=?\\\\a_{3}=a_{1}+(n-1)*r\\17=a_{1}+(3-1)*r\\17=a_{1}+(2*r)\\a_{1}=17-2r\\\\a_{20}=a_{1}+(20-1)*r\\a_{20}=a_{1}+(19*r)\\119=17-2r+19r\\119-17=19r-2r\\17r=102\\r=\frac{102}{17}\\razao=6\\\\a_{1}=17-(2*6)\\a_{1}=17-12\\a_{1}=5\\a_{12}=a_{1}+(12-1)*6\\a_{12}=5+(11*6)\\a_{12}=5+66\\\boxed {a_{12}=71}$

Explicação passo-a-passo:

Answer 2

resolução!

a20 = a3 + 17r

119 = 17 + 17r

119 - 17 = 17r

102 = 17r

r = 102/17

r = 6

a12 = a3 + 9r

a12 = 17 + 9 * 6

a12 = 17 + 54

a12 = 71

a16 da PA ( 3 , 8 , 13 .. )

r = a2 - a1

r = 8 - 3

r = 5

a16 = a1 + 15r

a16 = 3 + 15 * 5

a16 = 3 + 75

a16 = 78