Chapter 3.6, Problem 17PPS

To determine

To find the next three terms of the arithmetic sequence.

Answer to Problem 17PPS

  $3\frac{7}{12},4\frac{1}{3},5\frac{1}{12}$

Explanation of Solution

Given:

Sequence:

  $\frac{7}{12},1\frac{1}{3},2\frac{1}{12},2\frac{5}{6}\mathrm{...}$

Concept used:

An arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.

Calculation:

The common difference of the arithmetic sequence is

  $\begin{array}{c}1\frac{1}{3}-\frac{7}{12}=\frac{4}{3}-\frac{7}{12}\\ =\frac{16-7}{12}\\ =\frac{9}{12}\\ =\frac{3}{4}\end{array}$

Now, the next three consecutive numbers are:

  $\begin{array}{l}2\frac{5}{6}+\frac{3}{4}=\frac{17}{6}+\frac{3}{4}=\frac{34+9}{12}=\frac{43}{12}=3\frac{7}{12}\\ 3\frac{7}{12}+\frac{3}{4}=\frac{43}{12}+\frac{3}{4}=\frac{43+9}{12}=\frac{52}{12}=4\frac{4}{12}=4\frac{1}{3}\\ 4\frac{1}{3}+\frac{3}{4}=\frac{13}{3}+\frac{3}{4}=\frac{52+9}{12}=\frac{61}{12}=5\frac{1}{12}\end{array}$

Conclusion:

Therefore, the next three terms of the arithmetic sequence: $3\frac{7}{12},4\frac{1}{3},5\frac{1}{12}$ .