# How do you find the nth term rule for 1/2,1,3/2,2,...?

Nov 9, 2016

$n t h . = \frac{1}{2} n$

#### Explanation:

$\frac{1}{2} , 1 , \frac{3}{2} , 2 , \ldots$

firstly see what the terms increase by

difference $d = 1 - \frac{1}{2} = \frac{1}{2}$

so nth term is of the form

$n t h . = \frac{1}{2} n + c$

use the first term and $n = 1$ to evaluate $c$

$\frac{1}{2} = \frac{1}{2} \times 1 + c$

$\implies c = 0$

$\therefore n t h . = \frac{1}{2} n$

a quick mental check on the next few terms confirms the result.

Nov 9, 2016

${T}_{n} = \frac{1}{2} n$

#### Explanation:

The sequence might be easier to recognise as an AP if the terms are written as improper fractions:

${T}_{n} = \frac{1}{2} , \text{ "1," "1 1/2," " 2," " 2 1/2," " 3" } \ldots$

An even better way to write this is in halves:

${T}_{n} = \frac{1}{2} , \text{ "2/2," " 3/2, " "4/2," } \frac{5}{2.} . .$

$\text{ "n = 1, " " 2 ," " 3 , " "4 , " } 5 \ldots . .$

It is then immediately obvious that each term is just $\frac{n}{2}$

Using the formula/ general rule:

The nth term is written as ${T}_{n} = a + d \left(n - 1\right)$

All the values you need are available from the sequence.

$a = \frac{1}{2} , \text{ " d = 1/2" }$ substitute into ${T}_{n} = a + d \left(n - 1\right)$

${T}_{n} = \frac{1}{2} + \frac{1}{2} \left(n - 1\right) \text{ } \leftarrow$ multiply out and simplify

${T}_{n} = \frac{1}{2} + \frac{1}{2} n - \frac{1}{2}$

${T}_{n} = \frac{1}{2} n$