# How do you write the nth term rule for the sequence 5/2,11/6,7/6,1/2,-1/6,...?

Aug 22, 2016

${n}^{t h}$ term of the series is $\frac{19 - 4 n}{6}$

#### Explanation:

In $\left\{\frac{5}{2} , \frac{11}{6} , \frac{7}{6} , \frac{1}{2} , - \frac{1}{6.} . .\right\}$, let us make denominator common. As GCD is $6$, we can write the series as

$\left\{\frac{15}{6} , \frac{11}{6} , \frac{7}{6} , \frac{3}{6} , - \frac{1}{6.} . .\right\}$.

We observe that numerators form the series

$\left\{15 , 11 , 7 , 3 , - 1 , \ldots\right\}$ which form an arithmetic series with first term $a$ as $15$ and common difference $d$ between a term annd its preceding term is $12 - 15 = - 4$. Hence, ${n}^{t} h$ term of this series is $a + \left(n - 1\right) d$ i.e. 15+(n-1)×(-4)=15-4n+4=19-4n

Hence ${n}^{t h}$ term of the series is $\frac{19 - 4 n}{6}$