Arithmatic Progression for class 10
A progression is a special type of sequence for which it is possible to obtain a
formula for the nth term. The Arithmetic Progression is the most commonly used
sequence in maths with easy to understand formulas.
A mathematical sequence in which the difference between two consecutive terms
is always a constant and it is abbreviated as AP.
Where
a = First term
d = Common difference
n = number of terms
an = nth term
Example: Find the nth term of AP: 1, 2, 3, 4, 5…., an, if the number of terms are
15.
Solution: Given, AP: 1, 2, 3, 4, 5…., an
, n=15
By the formula we know, an = a+(n-1)d
First-term, a =1
Common difference, d=2-1 =1
Therefore, an = 1+(15-1)1 = 1+14 = 15
Sum of N Terms of AP
For any progression, the sum of n terms can be easily calculated. For an AP,
the sum of the first n terms can be calculated if the first term and the total terms are
known. The formula for the arithmetic progression sum is explained below:
Consider an AP consisting ―n‖ terms.
S = n/2[2a + (n − 1) × d]
This is the AP sum formula to find the sum of n terms in series.
Proof: Consider an AP consisting ―n‖ terms having the sequence a, a + d, a + 2d,
………….,a + (n – 1) × d
Sum of first n terms = a + (a + d) + (a + 2d) + ………. + [a + (n – 1) × d] ———
———-(i)
Writing the terms in reverse order,we have:
S = [a + (n – 1) × d] + [a + (n – 2) × d] + [a + (n – 3) × d] + ……. (a) ———–(ii)
Adding both the equations term wise, we have:
2S = [2a + (n – 1) × d] + [2a + (n – 1) × d] + [2a + (n – 1) × d] + …………. + [2a +
(n – 1) ×d] (n-terms)
2S = n × [2a + (n – 1) × d]
S = n/2[2a + (n − 1) × d]
Example: Let us take the example of adding natural numbers up to 15 numbers.
AP = 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15
Given, a = 1, d = 2-1 = 1 and an = 15
Now, by the formula we know;
S = n/2[2a + (n − 1) × d] = 15/2[2.1+(15-1).1]
S = 15/2[2+14] = 15/2 [16] = 15 x 8
A progression is a special type of sequence for which it is possible to obtain a
formula for the nth term. The Arithmetic Progression is the most commonly used
sequence in maths with easy to understand formulas.
A mathematical sequence in which the difference between two consecutive terms
is always a constant and it is abbreviated as AP.
Where
a = First term
d = Common difference
n = number of terms
an = nth term
Example: Find the nth term of AP: 1, 2, 3, 4, 5…., an, if the number of terms are
15.
Solution: Given, AP: 1, 2, 3, 4, 5…., an
, n=15
By the formula we know, an = a+(n-1)d
First-term, a =1
Common difference, d=2-1 =1
Therefore, an = 1+(15-1)1 = 1+14 = 15
Sum of N Terms of AP
For any progression, the sum of n terms can be easily calculated. For an AP,
the sum of the first n terms can be calculated if the first term and the total terms are
known. The formula for the arithmetic progression sum is explained below:
Consider an AP consisting ―n‖ terms.
S = n/2[2a + (n − 1) × d]
This is the AP sum formula to find the sum of n terms in series.
Proof: Consider an AP consisting ―n‖ terms having the sequence a, a + d, a + 2d,
………….,a + (n – 1) × d
Sum of first n terms = a + (a + d) + (a + 2d) + ………. + [a + (n – 1) × d] ———
———-(i)
Writing the terms in reverse order,we have:
S = [a + (n – 1) × d] + [a + (n – 2) × d] + [a + (n – 3) × d] + ……. (a) ———–(ii)
Adding both the equations term wise, we have:
2S = [2a + (n – 1) × d] + [2a + (n – 1) × d] + [2a + (n – 1) × d] + …………. + [2a +
(n – 1) ×d] (n-terms)
2S = n × [2a + (n – 1) × d]
S = n/2[2a + (n − 1) × d]
Example: Let us take the example of adding natural numbers up to 15 numbers.
AP = 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15
Given, a = 1, d = 2-1 = 1 and an = 15
Now, by the formula we know;
S = n/2[2a + (n − 1) × d] = 15/2[2.1+(15-1).1]
S = 15/2[2+14] = 15/2 [16] = 15 x 8
