Chapter 8.2, Problem 64E

To determine

To calculate: The value of x and the next term in the sequence.

Answer to Problem 64E

The value of x is $\frac{2}{3}$ and the terms are $\frac{7}{3},\frac{2}{3}\text{ and }-1$ .

Explanation of Solution

Given information:

The expression $3-x,x\text{ and }1-3x$ are the first three terms in an arithmetic sequence.

Formula used:

Arithmetic difference − the difference of consecutive term is constant. The constant difference is called the common difference and is denoted by d.

Therefore,

  $d=a_{n+1}-a_n;n\ge 1$ .

The $nth$ tem of an arithmetic sequence with first term $a_1$ and common difference $d$ is given by : $a_n=a_1+(n-1)d$ .

The sum of the first n terms of arithmetic series is

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

Calculation:

Consider the first three terms of the sequence.

Now,

  $\begin{array}{l}d=a_{n+1}-a_n\\ =a_2-a_1\\ =x-(3-x)\\ =x-3+x\\ =2x-3\end{array}$

The $nth$ term is $1-3x.$

Therefore,

  $\begin{array}{l}=a_n=a_1+(n-1)d\\ =1-3x=(3-x)+(3-1)(2x-3)\\ =1-3x=(3-x)+2(2x-3)\\ =1-3x=3-x+4x-6\\ =1-3x=3x-3\\ =1+3=3x+3x\\ =4=6x\\ =\frac{4}{6}=x\\ =\frac{2}{3}=x\end{array}$

Now, the sequence will be −

  $\begin{array}{l}=3-\left(\frac{2}{3}\right),\frac{2}{3}\text{ and }1-3\left(\frac{2}{3}\right)\\ =\left(\frac{9-2}{3}\right),\frac{2}{3}\text{ and }\left(1-2\right)\\ =\frac{7}{3},\frac{2}{3}\text{ and }-1.\end{array}$

Hence , the value of x is $\frac{2}{3}$ and the terms are $\frac{7}{3},\frac{2}{3}\text{ and }-1$ .