Chapter 8, Problem 26RE

a.

To determine

To find: the first 12 terms of the sequence.

Answer to Problem 26RE

  $\begin{array}{c}{a}_1=58.982\\ a_2=61.208\\ a_3=63.378\\ a_4=65.492\\ a_5=67.55\\ a_6=69.552\end{array}$

And

  $\begin{array}{c}{a}_7=71.498\\ a_8=73.388\\ a_9=75.222\\ a_{10}=77\\ a_{11}=78.722\\ a_{12}=80.388\end{array}$

Explanation of Solution

Given:

The average salaries from 2001 through 2012 can be approximated by is given by:

  $a_n=-0.028n^2+2.31n+56.7$ , $n=1,2,3,\mathrm{...},12$

To evaluate the first twelve terms of the sequence substitute $n=1,2,3,\mathrm{...},12$ in the above sequence.

  $\begin{array}{c}{a}_1=-0.028{\left(1\right)}^2+2.31\left(1\right)+56.7\\ =58.982\\ a_2=-0.028{\left(2\right)}^2+2.31\left(2\right)+56.7\\ =61.208\\ a_3=-0.028{\left(3\right)}^2+2.31\left(3\right)+56.7\\ =63.378\\ a_4=-0.028{\left(4\right)}^2+2.31\left(4\right)+56.7\\ =65.492\\ a_5=-0.028{\left(5\right)}^2+2.31\left(5\right)+56.7\\ =67.55\\ a_6=-0.028{\left(6\right)}^2+2.31\left(6\right)+56.7\\ =69.552\end{array}$

And

  $\begin{array}{c}{a}_7=-0.028{\left(7\right)}^2+2.31\left(7\right)+56.7\\ =71.498\\ a_8=-0.028{\left(8\right)}^2+2.31\left(8\right)+56.7\\ =73.388\\ a_9=-0.028{\left(9\right)}^2+2.31\left(9\right)+56.7\\ =75.222\\ a_{10}=-0.028{\left(10\right)}^2+2.31\left(10\right)+56.7\\ =77\\ a_{11}=-0.028{\left(11\right)}^2+2.31\left(11\right)+56.7\\ =78.722\\ a_{12}=-0.028{\left(12\right)}^2+2.31\left(12\right)+56.7\\ =80.388\end{array}$

b.

To determine

To graph: the first 12 terms of the given sequence.

Explanation of Solution

Given:

The average salaries from 2001 through 2012 can be approximated by is given by:

  $a_n=-0.028n^2+2.31n+56.7$ , $n=1,2,3,\mathrm{...},12$

Construct the graph of 12 terms of the sequence as shown:

  

c.

To determine

To predict: the average salaries for the year 2013 through 2019.

Answer to Problem 26RE

  $\begin{array}{c}{a}_{13}=81.998\\ a_{14}=83.552\\ a_{15}=85.05\\ a_{16}=86.492\\ a_{17}=87.878\\ a_{18}=89.208\\ a_{19}=90.482\end{array}$

Explanation of Solution

Given:

The average salaries from 2001 through 2012 can be approximated by is given by:

  $a_n=-0.028n^2+2.31n+56.7$ , $n=1,2,3,\mathrm{...},12$

To evaluate the next seven terms of the sequence substitute $n=13,14,15,16,17,18,19$ in the above sequence.

  $\begin{array}{c}{a}_{13}=-0.028{\left(13\right)}^2+2.31\left(13\right)+56.7\\ =81.998\\ a_{14}=-0.028{\left(14\right)}^2+2.31\left(14\right)+56.7\\ =83.552\\ a_{15}=-0.028{\left(15\right)}^2+2.31\left(15\right)+56.7\\ =85.05\\ a_{16}=-0.028{\left(16\right)}^2+2.31\left(16\right)+56.7\\ =86.492\\ a_{17}=-0.028{\left(17\right)}^2+2.31\left(17\right)+56.7\\ =87.878\\ a_{18}=-0.028{\left(18\right)}^2+2.31\left(18\right)+56.7\\ =89.208\\ a_{19}=-0.028{\left(19\right)}^2+2.31\left(19\right)+56.7\\ =90.482\end{array}$