Chapter 8.2, Problem 65E

To determine

To find:The10^th partial sum of given arithmetic sequence.

Answer to Problem 65E

The 10^th partial sum of the given arithmetic sequence is $620$ .

Explanation of Solution

Given information:

An arithmetic sequence is given as $8,20,32,44,\mathrm{...}$

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.

  $8,20,32,44,\mathrm{...}$

Now, first term of the finite sequence is $a_1=8$ .

Common difference is

  $\begin{array}{c}d=a_2-a_1\\ =20-8\\ =12\end{array}$

10^th term of the sequence can be found as

. $\begin{array}{c}{a}_n=a_1+\left(n-1\right)d\\ a_{10}=8+\left(10-1\right)12\\ =8+9\times 12\\ =8+108\\ =116\end{array}$

Number of term for the required sumis indicated as 10.

So, the 10^th partial sum is calculated as

  $\begin{array}{c}{S}_n=\frac{n}{2}\left(a_1+a_n\right)\\ S_{10}=\frac{10}{2}\left(8+116\right)\\ =5\times 124\\ =620\end{array}$

Thus, the 10^th partial sum of the given arithmetic sequence is $620$ .