Chapter 12.2, Problem 12AYU

In Problems 7-16, show that each sequence is arithmetic. Find the common difference and write out the first four terms.

$\left\{t_n\right\}=\left\{\frac{2}{3}+\frac{n}{4}\right\}$

To determine

To find: Each sequence is arithmetic and the common difference. Write out the first four terms $\left\{t_n\right\}=\left\{\frac{2}{3}+\frac{n}{4}\right\}$.

Answer to Problem 12AYU

$d=\frac{1}{4}$, $t_1=\frac{11}{12}$, $t_2=\frac{7}{6}$, $t_3=\frac{17}{12}$, $t_4=\frac{5}{3}$

Explanation of Solution

Given:

$\left\{t_n\right\}=\left\{\frac{2}{3}+\frac{n}{4}\right\}$

Formula used:

Calculation:

The $n^{\text{th}}$ and ${\left(n-1\right)}^{\text{th}}$ term of the sequence $\left\{t_n\right\}$ are

$t_n=\frac{2}{3}+\frac{1}{4}n\text{ and }{t}_{n-1}=\frac{2}{3}+\frac{1}{4}n-\frac{1}{4}=\frac{5}{12}+\frac{1}{4}n$

$d=t_n-t_{n-1}=\frac{2}{3}+\frac{1}{4}n-\frac{5}{12}-\frac{1}{4}n=\frac{3}{12}=\frac{1}{4}$   (a constant)

Since the difference of any two successive term is the constant, the given sequence is arithmetic and the common difference is $\frac{1}{4}$.

The first four terms are,

$t_1=\frac{2}{3}+\frac{1}{4}=\frac{11}{12}$

$t_2=\frac{2}{3}+\frac{2}{4}=\frac{14}{12}=\frac{7}{6}$

$t_3=\frac{2}{3}+\frac{3}{4}=\frac{17}{12}$

$t_4=\frac{2}{3}+\frac{4}{4}=\frac{20}{12}=\frac{5}{3}$