a geometric series has a sum of 9,837. the common ratio is 3 and the first term is 9. how many terms are in a series? answer

Respuesta :

[tex]S_n=9(3)^0+9(3)^1+\cdots+9(3)^{n-1}+9(3)^n[/tex]
[tex]3S_n=9(3)^1+9(3)^2+\cdots+9(3)^n+9(3)^{n+1}[/tex]

[tex]S_n-3S_n=9(3)^0-9(3)^{n+1}[/tex]
[tex]-2S_n=9(1-3^{n+1})[/tex]
[tex]S_n=\dfrac92(3^{n+1}-1)[/tex]

We're given that

[tex]S_n=9837=\dfrac92(3^{n+1}-1)[/tex]
[tex]\implies2186=3^{n+1}-1[/tex]
[tex]\implies2187=3^{n+1}[/tex]
[tex]\implies\log_32187=\log_33^{n+1}=n+1[/tex]
[tex]\implies7=n+1[/tex]
[tex]\implies n=6[/tex]

so there are 6 terms in the series.