Question

determine quantos termos tem a pg(6,18,...,1458)

Answer 1

$PG~(6, 18, ..., 1458)\\\\\\ q= \frac{18}{6} =3\\ a_1=6\\ a_n=1458\\ n=?\\\\\\ a_n=a_1\times q^{n-1}\\\\\\ 1458=6\times 3^{n-1}\to\\\\ \dfrac{1458}{6} =3^{n-1}\to\\\\ 243=3^{n-1}\to\\\\ \not3^5=\not3^{n-1}\to\\\\ 5=n-1\to\\\\ 5+1=n\to\\\\ \large\boxed{n=6}\\\\\\\\ PG~(6, 18, 54, 162, 486, 1458)$