Topic 2: Arithmetic and Geometric Progression

Arithmetic Progression

Arithmetic Progression (AP) is difference between one term and the next is a constant, other words is, just add same value.

Example of Arithmetic Progression:

Example 1:

AP is to find common different is subtract:

   First (1) term – Second (2) term

   Second (2) term – First (1) term =1

So,       1+1 = 2

            2+1 = 3

            3+1 = 4

            4+1 = 5

The constant by adding (+1) to next no.

Example 2:

Find the sum of the first 10 terms for the series that we met above: $\displaystyle{4},{7},{10},{13},\ldots$

a= 4

$\displaystyle{d}={3}$

$\displaystyle{n}={10}$

$\displaystyle{S}_{{n}}=\frac{n}{{2}}{\left[{2}{a}_{{1}}+{\left({n}-{1}\right)}{d}\right]}=\frac{10}{{2}}{\left[{2}\times{4}+{9}\times{3}\right]}={175}$

$S$n =  2/n [2a+(n-1)d]

=10/2 [2×4+9×3]

=175

Example 3:

Find the sum of the first $\displaystyle{1000}$ odd numbers.

In this case, we have $\displaystyle{a}_{{1}}={1}$, $\displaystyle{d}={2}$ and $\displaystyle{n}={1000}.$

So, using the formula:

$\displaystyle{S}_{{n}}=\frac{n}{{2}}{\left[{2}{a}_{{1}}+{\left({n}-{1}\right)}{d}\right]}$

$S$n =  2/n [2a+(n-1)d]

the sum 

$\displaystyle{1}+{3}+{5}+\ldots$ for 1000 steps is given by:

$\displaystyle{S}_{{1000}}=\frac{1000}{{2}}{\left[{2}{\left({1}\right)}+{\left({1000}-{1}\right)}{\left({2}\right)}\right]}={1}\ {000}\ {000}$[2(1)+(1000−1)(2)]=1 000 000

AP have 2 types formula:

            1] to find any term

           n th term T_n
          = a + (n-1)d


           2] to find sum term

          S_n = n/2 [2a + (n-1)d] or S_n = n/2 [a + 1]

SEE THIS VIDEO! (learn something from this video).....

SOLVE THIS QUESTIONS?

1.      Find the number of terms in the following Arithmetic sequence

T_n = a + (n-1) d

a)     2,5,8…299

a=2

d=3

b)     2,4,6…246

a=2

d=2

2.      In an AP, the 24^th term is 122 and common different is 3, find the 5 series

122= a + (23)3

                                122=a +69…continue to answer…

3.      The first 20 terms of the of series 24⁺³,27,30,33..

S_n= n/2 [2a+(n-1)d]

a=24

d=3

Geometric Progression

Geometric Progression (GP) Sequence of numbers where the ratio between any two adjacent numbers is constant.

Example of Geometric Progression:

Example 1:

2,4,8,16,32,64,128,256…

This sequence has a factor of 2 between each number.

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

In General we write a Geometric Sequence like this:

{a, ar, ar², ar³, ... }

where:

• a is the first term, and
• r is the factor between the terms (called the "common ratio")

Example 2:

{1,2,4,8,...}

The sequence starts at 1 and doubles each time, so

• a=1 (the first term)
• r=2 (the "common ratio" between terms is a doubling)

And we get:

{a, ar, ar², ar³, ... }

= {1, 1×2, 1×2², 1×2³, ... }

= {1, 2, 4, 8, ... }

But be careful, r should not be 0:

               *When r=0, we get the sequence {a,0,0,...} which is not geometric

Example 3: 

x_n = ar^(n-1)

(We use "n-1" because ar⁰ is for the 1st term)

10, 30, 90, 270, 810, 2430, ...

This sequence has a factor of 3 between each number.

The values of a and r are:

• a = 10 (the first term)
• r = 3 (the "common ratio")

x_n = 10 × 3^(n-1)

So, the 4th term is:

x₄ = 10×3^(4-1) = 10×3³ = 10×27 = 270

And the 10th term is:

x₁₀ = 10×3^(10-1) = 10×3⁹ = 10×19683 = 196830

Formula of Geometric Progression:

        1) nth term T_n = ar^n-1

       2) S_n = a(1-rⁿ)/1-r or S_n = a(rⁿ – 1)/r-1

        3)  S_∞ = a/1-r

      

SOLVE THIS QUESTIONS?

1.      What is the eleventh term of Geometric Sequence

3,10,12,24…?

Tn=ar ^{n-1 (11-1=10)}

a=3

r=2

2.      What is the sum of the first eight terms of the geometric sequence 5, 15, 45, ... ?

S_n = a(1-rⁿ)/1-r

5, 15, 45, ... 
This sequence has a factor of 3 between each pair of numbers.

The values of a, r and n are:
• a = 5 (the first term) 
• r = 3 (the "common ratio")
• n = 8 (we want to sum the first 8 terms)