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Arithmetic progression
A progression in which the difference between any two consecutive terms his always as same fixed quantity, such a progression is called arithmetic progression or simply AP. And the fixed difference is called common difference.
If a is the first time and t is the common difference then general AP form is:
a, a+t, a+2t, a+3t ……..
Where we denote terms as follows:
$a_1$ = a
$a_2$ = a+t
$a_3$ = a+ 2t and so on.
General term $a_n$ = a+(n1)t . That is $a_n$ is the $n^th$ term with a being the first term and t is the common difference.
Examples of AP

1,2,3,4,5,6,7……..

1,3,5,7,9….

9,6,3,0,3….

1,11,21,31,….
Sum of first n terms of AP
statment: Let $S_n$ be the sum of first n terms of AP $a_1,a_2,a_3,…….a_n$ then $S_n$ = $\frac{n}{2}[2a+(n1)t]$ = $\frac{n}{2}[a_1+a_n]$
Proof:
Given the AP, $a_1,a_2,a_3,…….a_n$
$$S_n = a_1 + a_2 + a_3 + …… + a_n$$
$$S_n = a + (a+t) + (a+2t) + ….. + [a+ (n1)t]$$
$$S_n = \sum_{r=1}^n[a+(n1)t]$$
$$S_n=\sum_{r=1}^{n} a + \sum_{r=1}^{n}(n1)t $$
As a and t are constants we can take them out of summation
$$S_n = a \sum_{r=1}^{n} 1 + t \sum_{r=1}^{n} (n1)$$
$$Here \sum_{r=1}^{n} 1 = n \ because\ 1\ is\ added\ n\ times$$
$$S_n = a n + t [\sum_{r=1}^{n} n  \sum_{r=1}^{n} 1]$$
$$And\ we\ know\ that\ sum\ of\ first\ n\ natural\ numbers\ i.e. \sum_{r=1}^{n} n = \frac{n(n+1)}{2}$$
$$S_n = a n + t [ \frac{n(n+1)}{2}  n]$$
$$S_n = n [ a + t (\frac{(n+1)}{2} 1)]$$
$$S_n = n [ a + t \frac{(n1)}{2} ]$$
$$S_n = \frac{n}{2} [ 2a + t (n1)]$$
Points to be noted
 If every term in an AP is multiplied by a constant k then new series will also be in AP with common difference kt
 If every term in an AP is added with k then the new series will be in AP with common difference (k+t)
Arithmetic Mean (AM)
Definition: If $a_1, a_2, a_3, …..a_n$ are real numbers then $\frac{a_1 + a_2 + a_3 + ….. + a_n}{n}$ is called arithmetic mean of $a_1, a_2, a_3, …..a_n$.
In such AM we can observe something interesting. AM of any two numbers make another AP
i.e. if A is the AM of a and b then a, A, b are in AP
If b is the AM of a and c then a, b, c are in AP
Proof:
Given b is the arithematic mean of a and c that means
b = $\frac{a+c}{2}$
for a, b ,c to be in AP the difference between adjacent terms must be equal i.e.
ba = c  b
as b = $\frac{a+c}{2}$
ba = $\frac{a+c}{2}$  a = $\frac{ca}{2}$
and cb = c  $\frac{a+c}{2}$ = $\frac{ca}{2}$
we got ba = cb
therefore a,b,c are in AP.
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