# Week 2 : Geometric Sequence

Geometric Sequences
In a Geometric Sequence each term is found by multiplying the previous term by a constant.

Example : 2,6,18,54, 163,486,1458,…

This sequence has a factor of 2 between each number

Each term is found by multiplying the previous term by 3 ( except the first term : 2 )

Infinite Gemetric Series :

In Infinite Geoemtric Series, when : r>1 or r<-1 it will be converges Let's see my example, and see what happends : 1,1/3,1/9,1/27,.. We have: a = 1 (the first term) r = 1/3 (halves each time) And so use this formula : $S_ \infty$ $S_ \infty$=1/1-1/3 $S_ \infty$=2/3 (0,666666..)

# 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)

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

# Arithmetic Sequences

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

My Arithmetic Sequences : 11,2,-7,-16,-25,-34

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

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

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