Answer:
[tex]\sum_{n=1}^{14}(3n-28)[/tex]
Step-by-step explanation:
The given series is
[tex]-25-22-19+...[/tex]
The the first term of the sequence is;
[tex]a_1=-25[/tex]
The constant difference is;
[tex]d=-22--25[/tex]
[tex]d=3[/tex]
The nth term of the series is
[tex]a_n=a_1+d(n-1)[/tex]
[tex]a_n=-25+3(n-1)[/tex]
[tex]a_n=-25+3n-3[/tex]
[tex]a_n=3n-28[/tex]
Using the summation notation, we can write the series as;
[tex]\sum_{n=1}^{14}(3n-28)[/tex]
