Answer:
[tex]\sum_{n=1}^{25}(3n-2)=925][/tex]
Step-by-step explanation:
The given series is [tex]\sum_{n=1}^{25}(3n-2)[/tex]
The first term of this series is
[tex]a_1=3(1)-2[/tex]
[tex]a_1=3-2[/tex]
[tex]a_1=1[/tex]
The second term is
[tex]a_2=3(2)-2[/tex]
[tex]a_2=6-2[/tex]
[tex]a_2=4[/tex]
The common difference is
[tex]d=4-1=3[/tex]
The sum of the first n-terms is given by;
[tex]S_n=\frac{n}{2}[2a_1+d(n-1)][/tex]
The sum of the first 25 terms of the series is
[tex]S_{25}=\frac{25}{2}[2(1)+3(25-1)][/tex]
[tex]S_{25}=\frac{25}{2}[2+3(24)][/tex]
[tex]S_{25}=\frac{25}{2}(74)][/tex]
[tex]S_{25}=925][/tex]
