In this problem, we have an arithmetic sequence with:
• first term a_1 = -22,
,
• common difference r = 5.
The terms of the arithmetic sequence are given by the following relation:
[tex]a_n=a_1+r\cdot(n-1)\text{.}[/tex]
Replacing the values a_1 = -22 and r = 5, we have:
[tex]a_n=-22+5\cdot(n-1)=-22+5n-5=5n-27.[/tex]
We must compute the sum of the first 30 terms of the sequence.
The sum of the first N terms of a sequence is:
[tex]\begin{gathered} S=\sum ^N_{n\mathop=1}a_n=\sum ^N_{n\mathop{=}1}(5n-27) \\ =5\cdot\sum ^N_{n\mathop{=}1}n-27\cdot\sum ^N_{n\mathop{=}1}1 \\ =5\cdot\frac{N\cdot(N+1)}{2}-27\cdot N. \end{gathered}[/tex]
Where we have used the relations:
[tex]\begin{gathered} \sum ^N_{n\mathop{=}1}n=\frac{N\cdot(N+1)}{2}, \\ \sum ^N_{n\mathop{=}1}1=N\text{.} \end{gathered}[/tex]
Replacing the value N = 30 in the formula for the sum S, we get:
[tex]S=5\cdot\frac{30\cdot31}{2}-27\cdot30=1515.[/tex]
Answer
sum = 1515
