Chapter 8.2, Problem 57E

To determine

To find:The sum of the given finite arithmetic sequence.

Answer to Problem 57E

The sum of the given finite arithmetic sequence is $110$ .

Explanation of Solution

Given information:

A finite arithmetic sequence is given as $2+4+6+8+10+12+14+16+18+20$

Concept used:

An arithmetic sequence of n terms, has the form $a_1,a_2,a_3,\mathrm{...},a_n$ and should have a common difference between each two consecutive terms.

That is

  $a_2-a_1=a_3-a_2=a_4-a_3=\mathrm{...}=a_n-a_{n-1}$

Common difference can be defined by d .

  $a_2-a_1=a_3-a_2=a_4-a_3=\mathrm{...}=a_n-a_{n-1}=d$

nth term of the arithmetic sequence has the form

  $a_n=a_1+\left(n-1\right)d$

Where $a_1$ is the first term of the sequence, and d is the common difference.

Sum of an arithmetic finite sequence has the form

  $S_n=\frac{n}{2}\left(a_1+a_n\right)$

Here, n is number of terms, $a_1$ is the first term of the sequence, and $a_n$ is the last term of the sequence.

Calculation:

Consider the given sequence.

  $2+4+6+8+10+12+14+16+18+20$

Now, first term of the finite sequence is $a_1=2$ .

Last term of the sequence is $a_n=20$ .

Common difference is

  $\begin{array}{c}d=a_2-a_1\\ =4-2\\ =2\end{array}$

Number of term of the sequence can be found as

  $\begin{array}{c}{a}_n=a_1+\left(n-1\right)d\\ 20=2+\left(n-1\right)2\\ 20=2+2n-2\\ 20=2n\\ n=10\end{array}$

So, the sum of the finite sequence is calculated as

  $\begin{array}{c}{S}_n=\frac{n}{2}\left(a_1+a_n\right)\\ S_{10}=\frac{10}{2}\left(2+20\right)\\ =5\times 22\\ =110\end{array}$

Thus, the sum of the finite arithmetic sequence is $110$ .