Answer:
The sum of the arithmetic series
[tex]S_{n} = \frac{n}{2} (2 a + (n-1) d)[/tex]
Step-by-step explanation:
Explanation
Let a , a+d , a+2 d , ..........a+(n-1)d +....... is an arithmetic sequence
The sum of the sequence is called arithmetic series
The [tex]n^{th}[/tex] term of the sequence
[tex]t_{n} = a + (n-1) d[/tex]
The sum of the arithmetic series
[tex]S_{n} = \frac{n}{2} (2 a + (n-1) d)[/tex]
Here 'a' is the first term of the sequence
and 'd' be the difference between two values
sum of first term
put n=1 ⇒ [tex]S_{1} = a[/tex]
Put n =2 ⇒ [tex]S_{2} = \frac{2}{2} (2 a + (2-1)d)[/tex]
[tex]S_{2} = 2 a + d[/tex]
......and so on
The sum of the arithmetic series
[tex]S_{n} = \frac{n}{2} (2 a + (n-1) d)[/tex]
