# Week1 : Arithmetic Sequences

$t_n$ $t_{50}$ $S_n=\frac{n}{2}(t_1+t_n)$

In an Arithmetic Sequence the difference between one term and the next is a constant.

In General we could write an arithmetic sequence like this:

{at, t+d, t+2d, t+3d, … }

where:

t is the first term, and
d is the difference between the terms (called the “common difference”)
Example : My Arithmetic Sequences : 11,2,-7,-16,-25,-34

Know :
– t = 11 (the first term)
– d = -9 (the “common difference” between terms)

We can write an Arithmetic Sequence as a formula :
$t_n$= t1+(n-1).d

Using the Arithmetic Sequence formula:

$tn$=t1+(n-1).d
$tn$=11+(n-1).(-9)
-34=11+9-9n
-34=20-9n
n=(6)

Example 2 : I want to determine t50
I know : t1= 11
n=50
n-1=50-1=49
Using the Arithmetic Sequence Rule :

$t_{50}$= t1+49.d
$t_{50}$= 11+49.(-9)
$t_{50}$= -430

To sum up the terms of this arithmetic sequence:

t1 + (t2+d) + (t3+2d) + (t4+3d) + …

Use this formula: $S_n=\frac{n}{2}(t_1+t_n)$

Example: Determine the sum of this Arithmetic Sequence : 11,2,-7,-16,-25,-34

The values of t,d and n are:

t = 11 (the first term)
d = -9 (the “common difference” between terms)
n (how many terms to add up)

$S_n=\frac{n}{2}(t_1+t_n)$
$S_n=\frac{6}{2}(11+-34)$
$S_n=(3.-23)$
$S_n=(-69)$