Question

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

$\left\{t_n\right\}=\left\{\frac{1}{2}-\frac{1}{3}n\right\}$

Short answer

The solution is $d=-\frac{1}{3}$; first four terms$=\frac{1}{6},-\frac{1}{6},-\frac{1}{2},-\frac{5}{6}$.

Step-by-step solution

Step 1. Given information 

$\left\{t_n\right\}=\left\{\frac{1}{2}-\frac{1}{3}n\right\}$

Step 2. Calculation 

The first term is$t_1=\frac{1}{2}-\frac{1}{3}·1=\frac{1}{6}$.

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

$t_n=\frac{1}{2}-\frac{1}{3}n$ and $t_{n-1}=\frac{1}{2}-\frac{1}{3}\left(n-1\right)=\frac{1}{2}-\frac{1}{3}n+\frac{1}{3}=\frac{5}{6}-\frac{1}{3}n$

The common difference is -

$d=t_n-t_{n-1}=\left(\frac{1}{2}-\frac{1}{3}n\right)-\left(\frac{5}{6}-\frac{1}{3}n\right)=-\frac{1}{3}$

Since the difference of any two successive terms is the constant $-\frac{1}{3}$, the sequence $\left\{t_n\right\}$is arithmetic.

The first four terms are$\frac{1}{6},-\frac{1}{6},-\frac{1}{2},-\frac{5}{6}$.