Chapter 12.1, Problem 41E

To determine

To calculate: The sequence which has 2 arithmetic means between $\sqrt{3}\text{ and 12}$ .

Answer to Problem 41E

The sequenceis $\sqrt{3},\frac{12+2\sqrt{3}}{3},\frac{24+\sqrt{3}}{3},12$ .

Explanation of Solution

Given information:

The sequence which has first term as $\sqrt{3}$ and last term as 12.

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 $\sqrt{3}$ and last term as 12.

According to question sequence will be of the form $\sqrt{3},a_2,a_3,12$ .

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, $a_1=\sqrt{3}$ , $n=4$ and $a_4=12$

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

Apply it for the $4\text{th}$ term,

  $\begin{array}{c}{a}_4=a_1+\left(4-1\right)d\\ 12=\sqrt{3}+3d\\ 12-\sqrt{3}=3d\\ d=\frac{12-\sqrt{3}}{3}\end{array}$

Therefore, common difference is $\frac{12-\sqrt{3}}{3}$ .

Now, second term is,

  $\begin{array}{c}{a}_2=a_1+d\\ =\sqrt{3}+\frac{12-\sqrt{3}}{3}\\ =\frac{12+2\sqrt{3}}{3}\end{array}$

And, third term is,

  $\begin{array}{c}{a}_3=a_2+d\\ =\frac{12+2\sqrt{3}}{3}+\frac{12-\sqrt{3}}{3}\\ =\frac{24+\sqrt{3}}{3}\end{array}$

Therefore, two arithmetic means are $\frac{12+2\sqrt{3}}{3}$ and $\frac{24+\sqrt{3}}{3}$ .

Thus, the sequenceis $\sqrt{3},\frac{12+2\sqrt{3}}{3},\frac{24+\sqrt{3}}{3},12$ .