Chapter 4.6, Problem 38E

To determine

To write: An equation for the $n\mathrm{th}$ term of the arithmetic sequence. Then find $a_{10}$ .

Answer to Problem 38E

  $a_n=\frac{2}{7}+\frac{1}{7}n$ .

  $a_{10}=\frac{12}{7}$ .

Explanation of Solution

Given information: Given sequence is $\frac{3}{7},\frac{4}{7},\frac{5}{7},\frac{6}{7}\mathrm{....}$ .

Calculation: Given sequence is $\frac{3}{7},\frac{4}{7},\frac{5}{7},\frac{6}{7}\mathrm{....}$ .

Common difference $=a_2-a_1=\frac{4}{7}-\frac{3}{7}=\frac{1}{7}$ .

  $n\mathrm{th}$ Term of arithmetic sequence is $a_n=a+(n-1)d$

Here $a=\frac{3}{7}\mathrm{and}d=\frac{1}{7}$ put this value in $a_n$

  $\begin{array}{l}{a}_n=a+(n-1)d\\ a_n=\frac{3}{7}+(n-1)\frac{1}{7}\\ a_n=\frac{3}{7}+\frac{1}{7}n-\frac{1}{7}\\ a_n=\frac{2}{7}+\frac{1}{7}n\end{array}$

Here $n=10$ put this value in $a_n$ .

  $\begin{array}{l}{a}_{10}=\frac{2}{7}+\frac{1}{7}(10)\\ a_{10}=\frac{2}{7}+\frac{10}{7}\\ a_{10}=\frac{12}{7}\end{array}$

Therefore $a_{10}=\frac{12}{7}$ .