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Arithmetic Progression Class 10  Ch5
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ARITHMETIC PROGRESSION
ARITHMETIC PROGRESSION
Class 10 Chapter 5 & Class 11 Chapter 9

Example: 2, 4, 6, 10, 12, 14, ................
Series :
If the terms of a sequence are connected by plus (or minus) sign, is called a series.Example: 2 + 4 + 6 + 10 + 12 + 14 + ................
Progression:
A sequence following some definite rule is called a progression.Arithmetic Progression
A sequence is called arithmetic progression if the difference of a term and its previous term is always same.
A sequence a_{1}, a_{2},
a_{3 },_{ }a_{4}, _{ }a_{5} ………, a_{n }, a_{n+1
}……… is called arithmetic progression if a_{n+1} = a_{n} + d, where a_{1} is called first term and d is called the common difference. 
3, 6, 9, 12, 15, ............
6  3 = 3, 9  6 = 3, 12  9 = 3, ..........t_{1}
Here we find that difference of any two consecutive terms is always remain same (3).
When difference of any two consecutive terms of a sequence remain same throughout the sequence, then that sequence is said to be in AP.
General Arithmetic Progression:
a , a + d, a + 2d, a + 3d, a + 4d +.............+ a + (n  1)d
Where
First Term(t_{1}) = a
Second Term(t_{2}) = a + d
Third Term(t_{3}) = a + 2d
..............................
.................................
Last term(t_{n}) = a + (n1)d
n^{th} term of sequence
is t_{n} = a + (n1)d Where common difference “d” is
given by: d = a_{n} – a_{n1} n^{th} terms of an AP from the end of the sequence is: l – (n1)d, where l is the last term of
the sequence. 
For example : 5, 10, 15, ............ 65.
Infinite Arithmetic Progression: When number of terms of an AP sequence are uncountable then AP is called infinite AP.
For example : 5, 10, 15, ...............
Four terms in AP can be taken as : a  3d, a  d, a + d, a + 3d
Five terms in AP can be taken as : a  2d , a  d , a , a + d , a + 2d
Six terms in AP can be taken as : a  5d, a  3d, a  d, a + d, a + 3d, a + 5d
Arithmetic Progression By Story Telling
Teacher can introduce this story just before introducing the topic Sum of n terms of an Arithmetic Progression.
LEARNING OUTCOMES
This story can enhance students' problemsolving skills, critical thinking abilities, and appreciation for mathematics.
It can also contribute to their understanding of patterns, sequences, and the historical development of mathematical concepts.
STORY
Once there was a young Boy of just 10 years old. Once, his elementary school teacher asked the class to add up the numbers from 1 to 100. The teacher intended to keep the students busy for a while, but this Boy quickly came up with the correct answer, 5050, much to the surprise of his teacher and classmates.
Rather than adding the numbers one by one, the boy noticed a pattern. He realized that if you pair the numbers at each end of the sequence, you get pairs that add up to the same sum. For example, the first number (1) and the last number (100) add up to 101, the second number (2) and the secondtolast number (99) add up to 101, and so on.
Since there are 100 numbers in the sequence and each pair adds up to 101, boy deduced that the total sum would be 101 multiplied by the number of pairs, which is 50. Thus, he quickly arrived at the sum of 5050.
This realization led that Boy to develop a general formula for finding the sum of an arithmetic series, where the series is a sequence of numbers with a common difference between them. The formula is given by:
Sum = (n/2) × (first term + last term),
where "n" represents the number of terms in the series.
The Boy, who deduced this formula was Carl Friedrich Gauss, a German mathematician and physicist who lived from 1777 to 1855.
Gauss made numerous contributions to various fields of mathematics, and one of his most notable achievements was discovering the formula for the sum of an arithmetic series.
Here we explain the method used by Gauss at that time in simple way
He wrote the numbers as follows
S = 1 + 2 + 3 + ................... + 99 + 100
And then, reversed the numbers to write
S = 100 + 99 + ................. + 2 + 1
Then he add both the sequences
2S = (100 + 1) + (99 + 2) + (98 + 3) + ............... + (2 + 99) + (1 + 100)
2S = 101 + 101 + 101 + .................... + 101 + 101 to 100 times
2S = 101 X 100 ⇒
If we take 100 = n, then n + 1 = 101,
So we can derive the formula for the sum of first n natural number
This formula allows us to calculate the sum of any arithmetic
series efficiently, without having to add up each term individually.
S = a + (a + d) + (a + 2d)+...................... + a + (n  1)d
S = a + (n  1)d + [a + (n  2)d] +...............+ (a + d) + a
Adding these two we get
2S = [a + a + (n  1)d] + [(a + d) + a + (n  2)d] + ................ + [a + (n  1)d + a]
2S = [2a + (n  1)d] + [2a + (n  1)d] + ........................... n times
2S = n X [2a + (n 1)d]
Sum of the first n terms of AP is
Where "a" is the first term and "d" is the common difference of the given AP sequence.
Sum of the first n terms of an AP is
Where "a" is the first term and "l" is the last term of the given AP sequence.
Sum of the first n terms from the end of the AP sequence is given by
nth term of an AP sequence is given by
Sum of n even natural numbers is given by:
Sum of first n odd natural numbers is given by
NOTE: Next topic is for the students after 10th standard in CBSE Board.
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b  a = d and c  b = d
⇒ b  a = c  b
⇒2b = a + b
1) If a constant is added to or subtracted from each term of an A.P. , then the resulting sequence is also an A.P. , with the same common difference.
2) If each term of an A.P. is multiplied or divided by an nonzero constant k, then the resulting sequence is also an A.P. with common difference kd or d/k, where d is the common difference of the given A.P.
3)In a finite A.P. the sum of the terms equidistant from the beginning and end is always same and is equal to the sum of first and last term.
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