Question

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

Answer 1

`S_n=9(3)^0+9(3)^1+\cdots+9(3)^(n-1)+9(3)^n`

`3S_n=9(3)^1+9(3)^2+\cdots+9(3)^n+9(3)^(n+1)`


`S_n-3S_n=9(3)^0-9(3)^(n+1)`

`-2S_n=9(1-3^(n+1))`

`S_n=\frac92(3^(n+1)-1)`

We're given that


`S_n=9837=\frac92(3^(n+1)-1)`

`\implies2186=3^(n+1)-1`

`\implies2187=3^(n+1)`

`\implies\log_32187=\log_33^(n+1)=n+1`

`\implies7=n+1`

`\implies n=6`

so there are 6 terms in the series.