Chapter 9.4, Problem 36E

Solution

i.

To state: The value of d if the balances shown in the table below from an arithmetic sequence. The table shows December balance in a simple interest savings account each year from 2014 through 2018.

  

The resultant answer is 2016.

Given information:

The given table is:

  

Explanation:

The balance shows an arithmetic sequence and the difference of any two successive savings account balances will be the common ratio.

Then the common difference is:

  $\begin{array}{l}d=\$20016-\$18000,d=2016\\ d=\$22032-\$20016,d=2016\end{array}$

Therefore, the common difference d is equal to 2016.

ii.

To state: The formula for the balance in the account n year after December 2014.

The formula is $\$18000+2016n$ .

Given information:

The value of d from part (a) is 2016 and the given table is:

  

Explanation:

Consider the given table:

  

The value of d from part (a) is 2016 it means every year, the balance becomes 2016 more than the previous years’ balance.

Then the formula for the balance in the account n years after December 2014 will become:

  $\begin{array}{c}a+nd=\$18000+n\left(2016\right)\\ =\$18000+2016n\end{array}$

Therefore, the formula is $\$18000+2016n$ .

iii.

To calculate: The sum of the December balances from 2014 to2024, inclusive.

The answer is $\$308,880$ .

Given information:

The value of d from part (a) is 2016 and the given table is:

  

Formula used: The sum of a finite arithmetic sequence $\left\{a_1,a_2,\cdots a_n\right\}$ with common difference d is given by: $Sum=\frac{n}{2}\left(2a+\left(n-1\right)d\right)$

Calculation:

Consider the given table:

  

The total number of years from 2014 to 2024 inclusive is 11. Then the value n is 11.

Substitute the values in the sum formula $Sum=\frac{n}{2}\left(2a+\left(n-1\right)d\right)$ ,

  $\begin{array}{c}Sum=\frac{11}{2}\left(2\left(18000\right)+\left(11-1\right)2016\right)\\ =\frac{11}{2}\left(36000+20160\right)\\ =\frac{11}{2}\left(56160\right)\\ =308,880\end{array}$

Therefore, the sum is $\$308,880$ .