Arithmetic Progression

Arithmetic Progression (AP) is the addition of consecutive  terms.
Eg. 1+2+3......+100 , 2+6+10...+22

Consecutive means one after the other so there is an equal difference between each term

So, to add on to the example, the AP would be something like this:

a + (a+d) + (a+2d) + ... + (a+[n-1]d) ,

where a is the first term and d is the common difference.

The n -th term of an AP can be found by:

a_n=a+(n-1)d

The formula is quite self-explanatory.

The sum of the first n terms of an AP also can be found by applying the formula below:

\displaystyle{S_n=\frac{1}{2}n(a+a_n)}

In order to find the number of terms in an AP, we must apply the formula below:

\displaystyle{n=\frac{a_n-a}{d}+1}

Example 1

Find the value of \displaystyle{\sum_{n=1}^{100} n} .

Solution 1

Here I introduced the Sigma Notation. It basically represents the word ‘sum’.
The sum a_1+a_2+...+a_n can be written as \displaystyle{\sum^n_{r=1} a_r} . In general,

\displaystyle{\sum^n_{r=m} a_r = a_m+a_{m+1}+...+a_n} , where m \leq n and m,n \in Z .

So, in the question, it is asking us to find out the sum of 1+2+3+...+100 . By applying the sum of AP, we get:

\displaystyle{S_n=\frac{1}{2}(100)(1+100)=50\cdot101=5050} .

Random Questions to try out:

1) 1+2+3+...+99

2) 55+57+59...+201

More Chim stuff…

Sum of Squares: \displaystyle{1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}}

Sum of Cubes: \displaystyle{1^3+2^3+3^3+...+n^3=(1+2+3+...+n)^2=\frac{n^2(n+1)^2}{4}}

Sun of Triangle numbers: \displaystyle{1+3+6+...+\frac{n(n+1)}{2}=\frac{n(n+1)(n+2)}{6}}

All the proofs of the properties above requires knowledge of Mathematical Induction (MI) which is really hard to explain so we’ll leave it as such for now. Do go and read up on MI

More random questions:

3) ??? ( SMOPS 2010)

Here is a program for arithmetic progression just for fun.

Arithmetic Progression
First Number:
Difference:
Number of Progressions:

Thks,

etzhkysy