Chapter 8.2, Problem 29E

To determine

To find: an expression for the nth term of the arithmetic sequence for given sixth term and twelfth term.

Answer to Problem 29E

An expression for the nth term of the arithmetic sequence is $a_n=103-3n$ .

Explanation of Solution

Given information:

An sequence is given with sixth term $a_6=26$ and twelfth term $a_{12}=42$ .

Concept used:

An arithmetic sequence of n terms, has the form $a_1,a_2,a_3,\mathrm{...},a_n$ and should have a common difference between each two consecutive terms.

That is

  $a_2-a_1=a_3-a_2=a_4-a_3=\mathrm{...}=a_n-a_{n-1}$

Common difference can be defined by d .

  $a_2-a_1=a_3-a_2=a_4-a_3=\mathrm{...}=a_n-a_{n-1}=d$

nth term of the arithmetic sequence has the form

  $a_n=a_1+\left(n-1\right)d$

Where $a_1$ is the third term of the sequence, and d is the common difference.

Calculation:

Now, consider the third term and the sixth term.

  $a_3=94$ and $a_6=85$

Consider dbe the common difference and $a_1$ as first term.

Now, expression for sixth term will be

  $\begin{array}{c}{a}_6=a_1+\left(6-1\right)d\\ 85=a_1+5d\\ a_1+5d=85\end{array}$ …… (1)

Now, expression for third term will be

  $\begin{array}{c}{a}_3=a_1+\left(3-1\right)d\\ 94=a_1+2d\\ a_1+2d=94\end{array}$ …… (2)

Subtract equation (1) from equation (2) to find the common difference.

  $\begin{array}{c}{a}_1+5d-a_1-2d=85-94\\ 3d=-9\\ d=-3\end{array}$

Now, the first term can be found by substituting $-3$ for d in equation (2).

  $\begin{array}{c}{a}_1+2d=94\\ a_1+2\left(-3\right)=94\\ a_1-6=94\\ a_1=94+6\\ a_1=100\end{array}$

So, the nth term of the arithmetic sequence can be found as

  $\begin{array}{c}{a}_n=a_1+\left(n-1\right)d\\ =100+\left(n-1\right)\left(-3\right)\\ =100-3n+3\\ =103-3n\end{array}$

Thus, the expression for the nth term of the arithmetic sequence is $a_n=103-3n$ .