If the first three terms of an arithmetic sequence are $\displaystyle \frac{4}{5}, \frac{38}{35}, \frac{336}{245}$ how do you find the nth term of the sequence?? i figured the bottom to be $\displaystyle a_1 7^{n-1}$
Hello, Intsecxtanx!
The first three terms of an arithmetic sequence are: .$\displaystyle \frac{4}{5},\;\frac{38}{35},\;\frac{336}{245}$
How do you find the $\displaystyle n^{th}$ term of the sequence?
You should know that the $\displaystyle n^{th}$ term is: .$\displaystyle a_n \;=\;a_1 + (n-1)d$
. . where: .$\displaystyle a_1$ = first term, .$\displaystyle d$ = common difference.
The first term is: .$\displaystyle a_1 \:=\:\frac{4}{5}$
The common difference is: .$\displaystyle d \:=\:\frac{38}{35}-\frac{4}{5} \:=\:\frac{2}{7}$
Got it?