# Find the nth-term of the sequence whose first few terms are written out?

Apr 21, 2018

$f \left(n\right) = {a}_{z} {r}^{n - z}$

#### Explanation:

Okay, so first we have to figure out if this is an arithmetic or geometric sequence.

For an arithmetic sequence, you should have the ability to add a common difference $d$ to each term to go the next term. The way you find $d$ is by taking a term and subtracting the term before it. You could choose and consecutive pair from the set, but I will just choose the first two.

$d = \left(- \frac{1}{6}\right) - \left(- \frac{3}{2}\right)$
Then simplify. Remember the double negative turns into a positive. You will then get,

$d = \frac{4}{3}$.
Now we have to check if this difference is applicable to the entire set. I will try to add $d$ to the second term to get to the third term.
$\left(- \frac{1}{6}\right) + \left(\frac{4}{3}\right) =$ $\frac{7}{6}$

That is different than the third term, so we now know that we have a geometric sequence.

The process is similar, but now you want to find the common ratio, $r$. To do this we will take one term, and divide it by the term before it. Again, I will use the first and second term.

$r = \frac{- \frac{1}{6}}{- \frac{3}{2}} = \frac{1}{9}$

We know this is correct by process of elimination but if you want to check, then take one term multiply it by the common ratio, to see if you get the next term.

Moving onward, we can now create the formula to find the $n$ term of the sequence we also now know is $\text{geometric}$.

The basic form for a geometric sequence is
$f \left(n\right) = a {r}^{n}$

Although, that is assuming that $a$ is the first term. Since we do not know any of the term numbers, we cannot figure out exactly what a is (you may have looked over some earlier information because the term number of at least one term in the sequence is necessary to solve this).

But, with what we have, we can put in our known information, $r$. Making the final formula to finding n,

$f \left(n\right) = {a}_{z} {r}^{n - z}$

Where z is the term number.

Apr 21, 2018

${T}_{n} = \frac{2 n - 5}{{n}^{2} + n}$
Take, $n = 1 , 2 , 3 , 4 , 5 , \ldots$
$\implies {T}_{1} = - \frac{3}{2} , {T}_{2} = - \frac{1}{6} , {T}_{3} = \frac{1}{12} , \ldots e t c .$

#### Explanation:

The given sequence is :

$\left\{- \frac{3}{2} , - \frac{1}{6} , \frac{1}{12} , \frac{3}{20} , \frac{5}{30} , \ldots\right\}$

We can separate the given sequence in two parts.

$\left(1\right)$Numerator

$\left\{- 3 , - 1 , 1 , 3 , 5 , \ldots\right\} \to$ $\text{Arithmetic Progression}$

So, the nth term ${t}_{n} = a + \left(n - 1\right) d$

where, first term $a = - 3 \mathmr{and}$ common difference $d = 2$

$\therefore {t}_{n} = - 3 + \left(n - 1\right) \times 2 = - 3 + 2 n - 2$

i.e. color(blue)(t_n=2n-5

$\left(2\right)$ Denominator:

$\left\{2 , 6 , 12 , 20 , 30 , \ldots\right\}$

${a}_{\textcolor{red}{1}} = 2 = {\textcolor{red}{1}}^{2} + \textcolor{red}{1}$

${a}_{\textcolor{red}{2}} = 6 = {\textcolor{red}{2}}^{2} + \textcolor{red}{2}$

${a}_{\textcolor{red}{3}} = 12 = {\textcolor{red}{3}}^{2} + \textcolor{red}{3}$

${a}_{\textcolor{red}{4}} = 20 = {\textcolor{red}{4}}^{2} + \textcolor{red}{4}$

${a}_{\textcolor{red}{5}} = 30 = {\textcolor{red}{5}}^{2} + \textcolor{red}{5}$

....

${a}_{\textcolor{red}{n}} = \ldots = {\textcolor{red}{n}}^{2} + \textcolor{red}{n} \to n t h$ term.

Hence, for the given sequence,nth term :

${T}_{n} = {t}_{n} / {a}_{n}$

$\implies {T}_{n} = \frac{2 n - 5}{{n}^{2} + n}$