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If we talk about arithmetic progression, it maintains a sequence of numbers or a series of numbers with the same gap or skips between the alternate numbers, and the difference between them remains constant. Let’s give you a very simple yet informative example of arithmetic progression.
For example, the sequence 5, 7, 9, 11, 13, 15, … is a progression with a standard difference of two.
When there is a finite portion of a progression between the numbers, it is called a limited progression, and sometimes it is just called a progression. The sum of a finite progression is named an arithmetic series.
The AP formulae are listed below.
An AP’s common difference is d = a2 – a1.
An AP’s nth word is a = a + (n – 1)d.
Sn = n2(2a+(n-1)d) is the sum of an AP’s n phrases.
An arithmetic progression (AP) may be a sequence where the differences between every two consecutive terms are equivalent. There is an opportunity to derive a formula for the nth term in a progression. For instance, the sequence 2, 6, 10, 14, … is a progression (AP) because it follows a pattern where each number is obtained by adding 4 to its previous term. During this sequence, nth term = 4n-2. The terms of the sequence are often obtained by substituting n=1,2,3, … within the nth term. i.e.,
When n = 1, 4n-2 = 4(1)-2 = 4-2=2
When n = 2, 4n-2 = 4(2)-2 = 8-2=6
When n = 3, 4n-2 = 4(3)-2 = 12-2=10
But how can we find the nth term of a given sequence?
Example 1.
Calculate the sum of the subsequent Arithmetic series:
1 + 8 + 15 + 22 + 29 + 36 + ………………… to 17 terms
Step by step solution of the sum of the progression arithmetic series:
The first term of the given arithmetic series = 1
The second term of the given arithmetic series = 8
The third term of the given arithmetic series = 15
The fourth term of the given arithmetic series = 22
The fifth term of the given arithmetic series = 29
Now, Second term – First term = 8 – 1 = 7
Third term – Second term = 15 – 8 = 7
Fourth term – Third term = 22 – 15 = 7
Therefore, the common difference of the given arithmetic series is 7.
The number of terms of the given A. P. series (n) = 17
We know that the sum of first n terms of the Arithmetic Progress, whose first term = a and customary difference = d is
S =n2(2a+(n-1)d)
Therefore, the specified sum of the first 20 terms of the series = 17/2[2 ∙ 1 + (17 – 1) ∙ 7]
= 17/2[2 + 16 ∙ 7]
= 17/2[2 + 112]
= 17/2 × 114
= 17 × 57
= 969
Derivation of progression Formula
Arithmetic progression is a progression only when the difference between the series of numbers in the sequence is the same and of the same interval. When the difference obtained by continuously adding a value turns out to be the same, it is known as a common difference. So, to seek out the nth term of a progression, we must understand the basic an = a1 + (n – 1)d. a1 is the first term, a1 + d is the second term, the third term is a1 + 2d, and so on. To locate the sum of the arithmetic series to fill the values, Sn, we start with the primary term and successively add on the common difference.
Sn = a1 + (a1 + d) + (a1 + 2d) + … + [a1 + (n–1)d].
We can also start with the nth term and successively subtract the common difference, so,
Sn = an + (an – d) + (an – 2d) + … + [an – (n–1)d].
Thus, the sum of the arithmetic sequence might be found in either of the ways. However, on adding those two equations together, we get
Sn = a1 + (a1 + d) + (a1 + 2d) + … + [a1 + (n–1)d]
Sn = an + (an – d) + (an – 2d) + … + [an – (n–1)d]
2Sn = (a1 + an) + (a1 + an) + (a1 + an) + … + [a1 + an].
Notice all the d terms are added out. So,
2Sn = n (a1 + an)
Sn = [n(a1 + an)]/2
By substituting an = a1 + (n – 1)d into the last formula, we have
Sn = n/2 [a1 + a1 + (n – 1)d] … Simplifying
Sn = n/2 [2a1 + (n – 1)d].
These two formulas help us quickly seek out the sum of an arithmetic series.
Conclusion
Arithmetic progression is just a simple sequence in which the difference between consecutive terms is constant. In this article, we learnt about some formulas to calculate the sum of arithmetic progression and the nth term of it. We also derived these formulas and took some examples to strengthen our concepts.
