Question

numa PA determinar a39 em que a1=3 e r=4

Answer 1

Termo geral da P.A:
$a_{n}=a_{1}+(n-1).r$.

P.A = { 3, 7, 11, 15,.......,$a_{39}$,.....,$a_{n}$}

$a_{1}=3$;
$r=4$;
$n=39$;
$a_{39}=?$.


$a_{39}=3+(39-1).4$ =>
$a_{39}=3+38.4$ =>
$a_{39}=3+152$ =>
$a_{39}=155$.

Answer 2

$a39 = 3 + (39 - 1)4$
a39= 3+38×4
a39=3+152
a39=155