Chapter 3.5, Problem 16PPS

To determine

Find next three terms of the sequence.

Answer to Problem 16PPS

The next three terms are $2\frac{1}{3},4,4\frac{1}{3}$

Explanation of Solution

Given:

The sequence: $2\frac{1}{3},2\frac{2}{3},3,3\frac{1}{3}\mathrm{.......}$

Concept Used:

An arithmetic sequence is a list of numbers with a definite pattern. If you take any number in the sequence then subtract it by the previous one and the result is always the same or constant then it is an arithmetic sequence.

In an Arithmetic Sequence the difference between one term and the next is a constant.

Calculation:

The sequence: $2\frac{1}{3},2\frac{2}{3},3,3\frac{1}{3}\mathrm{.......}$

Each term in an arithmetic sequence can be expressed in terms of the first term $a_1$ and the common difference $d$ . Since each succeeding term is formulated from one or more previous terms, this is a recursive formula. First term $a_1=2\frac{1}{3}$ and the common difference: $d=a_2-a_1=2\frac{2}{3}-2\frac{1}{3}=\frac{1}{3}$

Terms	Symbol	In terms of $a_1$ and d	Numbers
Fifth term	$a_5$	$a_1+4d$	$2\frac{1}{3}+4·(\frac{1}{3})=2\frac{1}{3}+\frac{4}{3}=2\frac{1}{3}+1\frac{1}{3}=3\frac{2}{3}$
Sixth term	$a_6$	$a_1+5d$	$2\frac{1}{3}+5·(\frac{1}{3})=2\frac{1}{3}+\frac{5}{3}=2\frac{1}{3}+1\frac{2}{3}=4$
Seventh term	$a_7$	$a_1+6d$	$2\frac{1}{3}+6·(\frac{1}{3})=2\frac{1}{3}+\frac{6}{3}=2\frac{1}{3}+2=4\frac{1}{3}$

The next three terms of the sequence $2\frac{1}{3},2\frac{2}{3},3,3\frac{1}{3}\mathrm{.......}$ are $2\frac{1}{3},4,4\frac{1}{3}$

Thus, the next three terms are: $2\frac{1}{3},4,4\frac{1}{3}$