Chapter 9.2, Problem 40PPSE

Solution

a.

To determine $6^{th}$ term, $8^{th}$ term and ${20}^{th}$ term.

  $a_6=-34,a_8=-48,a_{20}=-132$

Given:

The common difference is given as:

  $d=-7$

Formula Used:

An explicit formula for the arithmetic series is given as:

  $a_n=a+\left(n-1\right)d$

Calculation:

Let assume that the first term of the sequence is $a_1=1$ .And the common difference is $-7.$ Using calculator, the arithmetic series is given as:

  $1,-6,-13,-20,-27,-34,-41,-48,-55,-62,-69,-76,-83,-90,-97,-104,-111,-118,-125,-132$

Hence, the $6^{th}$ term is $-34,$

  $8^{th}$ term is $-48$ and the ${20}^{th}$ term is $-132.$

b.

To explain how the part (a) relates to the explicit formula.

Substituting the values of $a,d,n$ in an explicit formula, find the values of the sixth, eighth and twentieth terms.

Given:

The explicit formula is given as:

  $a_n=a+\left(n-1\right)d$

Interpretation:

In the explicit formula substitute the values of $a,n\text{ and }d$ and then find the $6^{th},8^{th}\text{ and }{20}^{th}$ terms.

Calculating the $6^{th}$ term as:

  $\begin{array}{l}{a}_6=a+\left(6-1\right)d\\ \Rightarrow a_6=1+5\left(-7\right)\\ \Rightarrow a_6=1-35\\ \Rightarrow a_6=-34\end{array}$

Next, calculating the $8^{th}$ term as:

  $\begin{array}{l}{a}_8=a+\left(8-1\right)d\\ \Rightarrow a_8=1+7\left(-7\right)\\ \Rightarrow a_8=1-49\\ \Rightarrow a_8=-48\end{array}$

Next, calculating the ${20}^{th}$ term as:

  $\begin{array}{l}{a}_{20}=a+\left(20-1\right)d\\ \Rightarrow a_{20}=1+19\left(-7\right)\\ \Rightarrow a_{20}=1-133\\ \Rightarrow a_{20}=-132\end{array}$

Hence, the $6^{th}$ term is $-34,$

  $8^{th}$ term is $-48$ and the ${20}^{th}$ term is $-132.$

Hence, the answer of part (a) relates to the explicit formula.