Chapter 12.1, Problem 39E

To determine

To calculate: The sequence which has 1 arithmetic meanbetween $12\text{ and 21}$ .

Answer to Problem 39E

The sequenceis $12,16.5,21$ .

Explanation of Solution

Given information:

The sequence which has first term as 12 and last term as 21.

Formula used:

A sequence with first term as $a_1$ and next term written as sum of preceding term and common difference then it is called an arithmetic sequence. It is denoted as $a_1,a_1+d,a_1+2d+\dots$ .

The $n\text{th}$ term of the arithmetic sequence is $a_n=a_1+\left(n-1\right)d$ .

Calculation:

Consider the sequence which has first term as 12 and last term as 21.

According to question sequence will be of the form $12,a_2,21$ .

Recall that a sequence with first term as $a_1$ and next term written as sum of preceding term and common difference then it is called an arithmetic sequence. It is denoted as $a_1,a_1+d,a_1+2d+\dots$ .

Therefore, $a_1=12$ , $n=3$ and $a_3=21$

The $n\text{th}$ term of the arithmetic sequence is $a_n=a_1+\left(n-1\right)d$ .

Apply it for the $3\text{rd}$ term,

  $\begin{array}{c}{a}_3=a_1+\left(3-1\right)d\\ 21=12+2d\\ 2d=9\\ d=4.5\end{array}$

Therefore, common difference is $4.5$ .

Now, second term is,

  $\begin{array}{c}{a}_2=a_1+d\\ =12+4.5\\ =16.5\end{array}$

Therefore, one arithmetic meanis $16.5$ .

Thus, the sequenceis $12,16.5,21$ .