Arithmetic progression is calculated from the first term and common difference, the sum of the progression of the specified range is displayed.
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Sum of arithmetic progression
About the calculation of Sum of arithmetic progression
Enter the first term, the common difference, and the range of the progression you want to calculate and click the button "Calculate sum of arithmetic progression". The sum of the arithmetic progression in the specified range is calculated.
It also shows how to calculate the sum of an arithmetic sequence.
Please enter up to 15 digits for First term and Common difference, and up to 15 digits for nth term.
What is Arithmetic progression?
An arithmetic progression is a sequence in which the difference between adjacent terms is equal.
The first term in a sequence is called the initial term, and the difference between each adjacent term is called the common difference.
For example, the following sequence has a first term of 1 and a common difference of 2.
The common difference is 2, so the difference between all adjacent terms is 2.
How to calculate Sum of arithmetic progression
Sum of the first to nth terms
Calculates the sum from the first term to the nth term.
If the first term is a_{1}, the nth term is a_{n}, and the common difference is d, the sequence is as follows.
a_{1}, a_{1}+d, a_{1}+2d, a_{1}+3d ... a_{n}−d, a_{n}
If the sum of the first through nth terms of this sequence is Sn, then Sn can be expressed as follows:
S_{n} = a_{1} + (a_{1}+d) + (a_{1}+2d) + (a_{1}+3d) + ... + (a_{n}−d) + a_{n}
Also, if you consider this sequence with the order reversed, the sum will be the same since only the order has changed.
Therefore, Sn can also be expressed as follows:
S_{n} = a_{n} + (a_{n}−d) + (a_{n}−2d) + ... + (a_{1}+2d) + (a_{1}+d) + a_{1}
Add the left and right sides of these two equations together.
S_{n} | = | a_{1} | + | (a_{1}+d) | + | (a_{1}+2d) | + ... + | (a_{n}−d) | + | a_{n} | |
+ | S_{n} | = | a_{n} | + | (a_{n}−d) | + | (a_{n}−2d) | + ... + | (a_{1}+d) | + | a_{1} |
2S_{n} | = | (a_{1}+a_{n}) | + | (a_{1}+a_{n}) | + | (a_{1}+a_{n}) | + ... + | (a_{1}+a_{n}) | + | (a_{1}+a_{n}) |
On the right-hand side, all the numbers added together become (a_{1} + a_{n}), and since the number of items is n and can be expressed as (a_{1} + a_{n}) × n, we get 2S_{n} = n(a_{1} + a_{n}).
Dividing both sides of this equation by 2 gives us S_{n} = 12n(a_{1} + a_{n}).
Therefore
Also, the nth term is a_{1} +(n − 1)d, so by substituting a_{n} = a_{1} +(n − 1)d, we get
S_{n} = 12n(2a_{1} +(n − 1)d).
Therefore
Sum of nth through mth term
To calculate the sum from the nth term to the mth term, you similarly reverse the sequence and add both sides.
S_{m−n} | = | a_{n} | + | (a_{n}+d) | + | (a_{n}+2d) | + ... + | (a_{m}−d) | + | a_{m} | |
+ | S_{m−n} | = | a_{m} | + | (a_{m}−d) | + | (a_{m}−2d) | + ... + | (a_{n}+d) | + | a_{n} |
2S_{m−n} | = | (a_{n}+a_{m}) | + | (a_{n}+a_{m}) | + | (a_{n}+a_{m}) | + ... + | (a_{n}+a_{m}) | + | (a_{n}+a_{m}) |
On the right-hand side, all the numbers added are (a_{n}+a_{m}), and the number of items is (m−n+1), so it can be expressed as (a_{n}+a_{m}) × (m−n+1), and therefore 2S_{m−n} = (m−n+1) × (a_{n}+a_{m}).
Dividing both sides of this equation by 2 gives us S_{m−n} = 12(m−n+1)(a_{n}+a_{m}).
Therefore