Question

Quantos termos tem a PG (1,5,....125) ?

Answer 1

$\textsf{F\'ormula de recorr\^encia de uma P.G:}\\\\\\\mathsf{a_1\cdot q=a_2}\\\\\mathsf{1\cdot q=5}\\\\\mathsf{q=\dfrac{5}{1}=5}\\\\\\\mathsf{Teorema~do~termo~geral:~~\fbox{ \mathsf{a_n=a_1\cdot q^{n-1}} }}\\\\\\\mathsf{Perceba~que~se~substituirmos~a_n~por~125~(\'ultimo~termo~da~P.G)~n\'os}\\\textsf{iremos descobrir sua posi\c{c}\~ao na progress\~ao, logo, tamb\'em descobrir\underline{e}}\\\textsf{mos quantos termos h\'a nessa P.G.}$


$\mathsf{125=1\cdot 5^{n-1}}\\\\\\\mathsf{125=5^{n-1}~~~~(fatorando~125...)}\\\\\\\mathsf{5^3=5^{n-1}~~~~(igualando~as~pot\^encias...)}\\\\\\\mathsf{3=n-1}\\\\\\\fbox{ \mathsf{n=4} }~~~~\checkmark\\\\\\\textsf{Portanto, essa P.G apresenta 4 termos, visto que o \'ultimo termo da}\\\textsf{progress\~ao est\'a na posi\c{c}\~ao 4.}$

Answer 2

resolução!

q = a2/a1

q = 5/1

q = 5

an = a1 * q^n - 1

125 = 1 * 5^n - 1

125 = 5^n - 1

5^3 = 5^n - 1

n - 1 = 3

n = 3 + 1

n = 4

resposta : PG de 4 termos