Question

In the problem, show that each sequence is arithmetic. Find the common difference and write out the first four terms.

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

Short answer

The solution is $d=\frac{1}{4}$; first four terms$=\frac{11}{12},\frac{7}{6},\frac{17}{12},\frac{5}{3}$.

Step-by-step solution

Step 1. Given information 

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

Step 2. Calculation 

The first term is$t_1=\frac{2}{3}+\frac{1}{4}=\frac{11}{12}$

The $n$th term and the $\left(n-1\right)$st term of the sequence are-

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

The common difference is -

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

Since the difference of any two successive terms is the constant $\frac{1}{4}$, the sequence $\left\{t_n\right\}$is arithmetic.

The first four terms are$\frac{11}{12},\frac{7}{6},\frac{17}{12},\frac{5}{3}$.