The document discusses arithmetic progression (A.P.), which is a series where the difference between consecutive terms is constant. It provides examples like the summation of natural numbers from 1 to n. The key aspects are: 1) An A.P. has a first term, a common difference between terms, and a formula to calculate any term as the first term plus the product of the common difference and the term number minus 1. 2) The formula for summing the first n natural numbers is n(n+1)/2. 3) By analogy, the formula for summing an A.P. from its first term a to its nth term is n(2a+(n-1)d)/

1. ARITHMETRIC PROGRESSION
(A.P.)

2. Consider the following series:
∑=1+2+3+4+5+6+7+8+9+………….+n
This is the sum of ‘n’ natural numbers.
First term in the series is 1
Second term in the series is 2, which is 1+1
Third term in the series is 3, which is 1+2
Between two successive numbers, the common difference is 1(for instance,(2-1),(3-2),(4-
3)).
Now, let us generalize this series.
Let the first term of the series be ‘a’.
Let the second term be (a + d), where ‘d’ is the common difference.
Third term will be (a+2d), such that (difference between third and second terms is ‘d’’.
Fourth term will be (a+3d)
Likewise if we keep writing, we get the nth term in the series as a+(n-1)d .

3. Now, we can write the general arithmetic series with n terms as:
∑= a+(a+d)+(a+2d)+(a+3d)+………….+(a+(n-1)d).
The series above is called arithmetic series and the successive terms in the series are said to be
in arithmetic progression.
If we consider three successive terms of a series as a, b ,c respectively, then,
b is said to be the arithmetic mean of a and c.
That is, b=(a + c)
2
Let us see some more examples of arithmetic series

4. Odd numbers summation series
∑=1+3+5+7+9+……..+(2n-1)
Even numbers summation series
∑=2+4+6+8+10+……..+(2n)
Let us derive formula for arithmetic progression summation series.
Before that let us derive the formula for sum of ‘n’ natural numbers.
Let ‘S’ represent the series of sum of ‘n’ natural numbers.
S= 1+2+3+……+n ….(Eqn. i)
Writing R.H.S. in reverse order, we get,
S=n+(n-1)+(n-2)+……+1 ….(Eqn. ii)

5. Adding Eqn (i) & Eqn (ii),
2S= (n+1)+(n-1+2)+(n-2+3)+………+(n+1)
2S=n(n+1)
S= n(n+1)
2
Therefore, sum of ‘n’ natural numbers is given by n(n+1), where 1 is the first term and ‘n’ is the last term.
2
Likewise, if ‘a’ is the first term of the series and ‘l’ is the last term of the series, then,
Sum of the series with ‘n’ natural numbers is
S= n (a+l)
2

6. Using this relation, we will arrive at sum of series for arithmetic progression.
S= n(a+l)
2
a = first term of the series
l = last term of the series
For arithmetic series, the nth term is a+(n-1)d ( as discussed earlier)
S= n(a + a+(n-1)d)
2
S=n(2a+(n-1)d)
2