Question

Finding Terms of an Arithmetic Sequence Find the indicated term of the arithmetic sequence with the given description.

The fourteenth term is $\frac{2}{3}$,and the ninth term is $\frac{1}{4}.$Find the first term and the $n\text{th}$term.

Short answer

The first term is$-\frac{5}{12}$and the$n\text{th}$ term is$a_n=-\frac{5}{12}+\frac{1}{12}\left(n-1\right)$.

Step-by-step solution

Step 1. Given information.

The fourteenth term is $\frac{2}{3}.$

The ninth term is$\frac{1}{4}.$

Step 2. Write the concept.

The $n\text{th}$ term of an arithmetic sequence is:

$a_n=a+\left(n-1\right)d$

Where $a$ is the first term and $d$ is a common difference.

Step 3. Determine the two equations.

The fourteenth term is $a_{14}=\frac{2}{3}.$

$\begin{array}{c}a+\left(14-1\right)\left(d\right)=\frac{2}{3}\\ a+13d=\frac{2}{3}\mathrm{...}\text{\hspace{0.33em}(i)}\end{array}$

The ninth term is $a_9=\frac{1}{4}.$

$\begin{array}{c}a+\left(9-1\right)\left(d\right)=\frac{1}{4}\\ a+8d=\frac{1}{4}\mathrm{...}\text{\hspace{0.33em}(ii)}\end{array}$

Step 4. Determine the first term and common difference.

Subtract equation (ii) from (i).

$\begin{array}{c}a+13d-a-8d=\frac{2}{3}-\frac{1}{4}\\ 5d=\frac{8-3}{12}\\ d=\frac{5}{12\times 5}\\ d=\frac{1}{12}\end{array}$

Substituterole="math" localid="1648727826298" $d=\frac{1}{12}$in (ii).

$\begin{array}{c}a+8\left(\frac{1}{12}\right)=\frac{1}{4}\\ a+\frac{2}{3}=\frac{1}{4}\\ a=\frac{1}{4}-\frac{2}{3}\\ a=\frac{3-8}{12}\\ a=-\frac{5}{12}\end{array}$

Step 5. Determine the nth term.

The $n\text{th}$term of the sequence is:

$\begin{array}{c}{a}_n=a+\left(n-1\right)d\\ =\left(-\frac{5}{12}\right)+\left(n-1\right)\left(\frac{1}{12}\right)\\ =-\frac{5}{12}+\frac{1}{12}\left(n-1\right)\end{array}$