Given the arithmetic series below
[tex]6+8+10+12+\cdots,\text{ n = 8}[/tex]
To find the sum of the arithmetic series,
We find the last term, i.e 8th term of the series
The formula to find the last term/formula for the arithmetic progression is given below
[tex]U_n=a+(n-1)d[/tex]
The common difference, d, is
[tex]2nd\text{ term - 1st term = 8 - 6 = 2}[/tex]
Where,
[tex]n=8,d=2,a=6[/tex]
Substitute the values into the formula to find the 8th term
[tex]\begin{gathered} U_8=6+(8-1_{})(2)=6+(7)(2)=6+14=20 \\ U_8=20 \end{gathered}[/tex]
The formula to find the arithmetic series is
[tex]S_n=\frac{n}{2}(a_1+a_n)_{}[/tex]
Where
[tex]a_1=6,n=8,a_8=U_8=20[/tex]
Subtsitute the values into the formula to find the arithmetic series above
[tex]\begin{gathered} S_8=\frac{8}{2}(6+20)=4(26)=104 \\ S_8=104 \end{gathered}[/tex]
Hence, the sum of the arithmetic series is 104
