Question

quantos termos tem a PG an=1875, a1=3 e q=5

Answer 1

$a_n=1875 \\ a_1=3 \\ q=5 \\ \\ a_n=a_1.q^{n-1} \\ \\ 1875=3.5^{n-1} \\ \\ 1875\div3=5^{n-1} \\ \\ 625=5^{n-1} \\ \\ 5^4=5^{n-1} \\ \\ n-1=4 \\ \\ n=4+1 \\ \\ n=5\mapsto 5~~termos$

Answer 2

$a_{n}=a_{1} * q^{n-1}\\\\1875 = 3*5^{n-1} \\1875/3 =5^{n-1}\\625 = 5^{n-1}\\5^4 = 5^{n-1}\\\\ 4 = n - 1\\n = 4 + 1 \\\\ n = 5$


Resposta: 5 termos.