arithmetic sequence help

**Intsecxtanx** · November 21st 2009, 02:33 PM

If the first three terms of an arithmetic sequence are $\frac{4}{5}, \frac{38}{35}, \frac{336}{245}$ how do you find the nth term of the sequence?? i figured the bottom to be $a_1 7^{n-1}$

**mosta86** · November 21st 2009, 03:10 PM

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.

**Soroban** · November 21st 2009, 03:46 PM

Hello, Intsecxtanx!

The first three terms of an arithmetic sequence are: . $\frac{4}{5},\;\frac{38}{35},\;\frac{336}{245}$
How do you find the $n^{th}$ term of the sequence?

You should know that the $n^{th}$ term is: . $a_n \;=\;a_1 + (n-1)d$

. . where: . $a_1$ = first term, . $d$ = common difference.

The first term is: . $a_1 \:=\:\frac{4}{5}$

The common difference is: . $d \:=\:\frac{38}{35}-\frac{4}{5} \:=\:\frac{2}{7}$

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