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Arithmetic series

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Summary

Arithmetic series

​​In a nutshell

An arithmetic series finds the sum of a given number of terms in an arithmetic sequence. 



Calculating the sum of a sequence

To find the sum SnS_nSn​ of the first nnn terms of an arithmetic sequence, you can use a formula. The sum SnS_nSn​ is given using the first term, aaa, the value nnn, which is the position of the nthn^{\text{th}}nth​ term, and either the common difference, ddd, or the last term, lll.​

​Sn=n2[2a+(n−1)d]orSn=n2(a+l)\boxed {S_n = \dfrac {n}{2} [2a+(n-1)d] \quad or \quad S_n = \dfrac {n}{2} (a+l)}Sn​=2n​[2a+(n−1)d]orSn​=2n​(a+l)​​​


Deriving the formula

The terms in any given finite arithmetic sequence are as follows: 

a, a+d,a+2d,...,a+(n−2)d, a+(n−1)da, \ a+d, a+2d,... , a+(n-2)d, \ a +(n-1)da, a+d,a+2d,...,a+(n−2)d, a+(n−1)d​​


This means that the sum of a finite arithmetic sequence will be as follows:

Sn=[a]+[a+d]+[a+2d]+...+[a+(n−2)d]+[a+(n−1)d]S_n=[a]+[a+d]+ [a+2d]+... +[ a+(n-2)d]+ [a +(n-1)d]Sn​=[a]+[a+d]+[a+2d]+...+[a+(n−2)d]+[a+(n−1)d]​​​​


Adding two such sums together, expressing one of them in reverse, will give: 

Sn=[a]+[a+d]+...+[a+(n−2)d]+[a+(n−1)d]Sn=[a+(n−1)d]+[a+(n−2)d]+...+[a+d]+[a]2Sn=[2a+(n−1)d]+[2a+(n−1)d]+...+[2a+(n−1)d]+[2a+(n−1)d]\begin{aligned} S_n&= &[a]&+ &[a+d] &+... &+&[ a+(n-2)d]&+ &[a +(n-1)d] \\ S_n&= &[a+(n-1)d]&+&[a+(n-2)d] &+... &+&[ a+d]&+ &[a ] \\ \hline 2S_n &= &[2a+(n-1)d] &+ &[2a+(n-1)d] &+ ... &+ &[2a+(n-1)d] &+&[2a+(n-1)d] \end{aligned}Sn​Sn​2Sn​​===​[a][a+(n−1)d][2a+(n−1)d]​+++​[a+d][a+(n−2)d][2a+(n−1)d]​+...+...+...​+++​[a+(n−2)d][a+d][2a+(n−1)d]​+++​[a+(n−1)d][a][2a+(n−1)d]​​​​


Therefore, since nnn is the total number of terms in the calculation, this will give:

2Sn=n(2a+(n−1)d)2S_n = n(2a+(n-1)d)2Sn​=n(2a+(n−1)d)​​

Sn=n2(2a+(n−1)d)\boxed{S_n = \dfrac {n}2 (2a+(n-1)d)}Sn​=2n​(2a+(n−1)d)​​​


If nnn represents the total number of terms, then the last term is l=a+(n−1)dl = a+(n-1)dl=a+(n−1)d. Hence an alternative formula can be found:

Sn=n2(a+[a+(n−1)d])S_n = \dfrac {n}2 (a+[a+(n-1)d]) Sn​=2n​(a+[a+(n−1)d])​​

​Sn=n2(a+l)\boxed{S_n = \dfrac {n}2 (a+l) }Sn​=2n​(a+l)​​​


Example 1

Find the sum of the first 252525 terms of the arithmetic sequence with a first term of 888, and a common difference of 333. 


You can identify the first few terms of the sequence:

​8,11,14,17,...8,11,14,17,...8,11,14,17,...​​


To sum the first fifteen terms, without actually finding them and adding them, identify the formula to use:

Sn=n2[2a+(n−1)d]S_n = \dfrac{n}2 [2a + (n-1)d] Sn​=2n​[2a+(n−1)d]​​

​

Substitute values to find the sum: 

S25=252[2(8)+(25−1)3]S_{25} = \dfrac {25}{2} [2(8) + ( 25-1)3]S25​=225​[2(8)+(25−1)3]​​

S25=1100‾\underline{S_{25} = 1100}S25​=1100​​​


Example 2

Given that the sum of the first 151515 terms of an arithmetic sequence is 630630630 and the last term is 767676, what is the first term of this sequence?


Identify the relevant formula to use:

Sn=n2(a+l)S_n = \dfrac {n}{2} (a+l)Sn​=2n​(a+l)​​


Substitute the given values and solve for aaa:​

S15=152(a+76)S_{15} = \dfrac{15}{2}(a+76)S15​=215​(a+76)​​

​630=152(a+76)630 = \dfrac{15}{2}(a+76)630=215​(a+76)​​

​84=a+7684 = a + 7684=a+76

​a=8‾\underline{a=8}a=8​​​



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FAQs - Frequently Asked Questions

What is an arithmetic series based on?

An arithmetic sequence.

What is the formula for an arithmetic series?

S_n = (n/2) (2a+(n-1)d) or S_n = (n/2) (a+l) where a is the first term of a sequence, n is the position in a sequence, and l is the last term.

What is an arithmetic series?

An arithmetic series finds the sum of a given number of terms in an arithmetic sequence.

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