Question

Hello! I need help solving and answering this practice problem. Having trouble with it.

Answer 1

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:

`a_n=a_1+r\cdot(n-1)\text{.}`

Replacing the values a_1 = -22 and r = 5, we have:

`a_n=-22+5\cdot(n-1)=-22+5n-5=5n-27.`

We must compute the sum of the first 30 terms of the sequence.

The sum of the first N terms of a sequence is:

`\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(N\cdot(N+1))/(2)-27\cdot N. \end{gathered}`

Where we have used the relations:

`\begin{gathered} \sum ^N_{n\mathop{=}1}n=(N\cdot(N+1))/(2), \\ \sum ^N_{n\mathop{=}1}1=N\text{.} \end{gathered}`

Replacing the value N = 30 in the formula for the sum S, we get:

`S=5\cdot(30\cdot31)/(2)-27\cdot30=1515.`

Answer

sum = 1515