## Precalculus (10th Edition)

$a_n=\dfrac{4-n}{3}$ $a_{51}=-\dfrac{47}{3}$
We know that the $n^{th}$ term of an aritihmetic sequence is given by the formula $a_n=a_1+(n-1)d$ where $d$=common difference and $a_1$= the first term Hence, here we have $a_n=1+(n-1)\cdot\left(-\dfrac{1}{3}\right)\\a_n=1+\left(-\dfrac{n}{3}\right)+\dfrac{1}{3}\\a_n=\dfrac{4}{3}-\dfrac{n}{3}\\a_n=\dfrac{4-n}{3}$ Therefore, $a_{51}=\dfrac{4-51}{3}\\a_{51}=-\dfrac{47}{3}$