# How do you write the rule for the nth term given -1/2,-1/4,-1/6,-1/8,...?

Jul 29, 2016

${n}^{t h}$ of the sequence $\left\{- \frac{1}{2} , - \frac{1}{4} , - \frac{1}{6} , - \frac{1}{8} , \ldots \ldots \ldots .\right\}$ is $- \frac{1}{2 n}$

#### Explanation:

It is observed in the sequence that while numerator is constant at $- 1$, the difference between the denominator of a term of the sequence to the denominator of its preceding term is constant at $2$, as denominators are $\left\{2 , 4 , 6 , 8 , \ldots \ldots \ldots .\right\}$.

Clearly, it is not the sequence but its denominators, which are in arithmetic sequence $\left\{2 , 4 , 6 , 8 , \ldots \ldots \ldots .\right\}$, with first term as $2$ and common difference at $2$.

As the ${n}^{t h}$ of a sequence whose first term is $a$ and common difference is $d$ is $a + \left(n - 1\right) d$, ${n}^{t h}$ of the sequence $\left\{2 , 4 , 6 , 8 , \ldots \ldots \ldots .\right\}$ is

$2 + \left(n - 1\right) \times 2 = 2 + 2 n - 2 = 2 n$.

Hence ${n}^{t h}$ of the sequence $\left\{- \frac{1}{2} , - \frac{1}{4} , - \frac{1}{6} , - \frac{1}{8} , \ldots \ldots \ldots .\right\}$ is $- \frac{1}{2 n}$