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 $\displaystyle a_1 7^{n-1}$

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- Nov 21st 2009, 02:33 PMIntsecxtanxarithmetic sequence help
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 $\displaystyle a_1 7^{n-1}$

- Nov 21st 2009, 03:10 PMmosta86
assume U0=4/5 U1=38/35 U2=336/245

Un=U0+nr

r=U1-U0=U2-U1= 10/35

=> Un= 4/5 + n ( 10/35 ) general term. - Nov 21st 2009, 03:46 PMSoroban
Hello, Intsecxtanx!

Quote:

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?