Chapter 9.4, Problem 35E

Solution

i.

To state: The value of r if the balances shown in the table below from a geometric sequence. The table shows December balance in a fixed-rate compound savings account each year from 2014 through 2018.

  

The resultant answer is 1.1.

Given information:

The given table is:

  

Explanation:

The balance shows a geometric sequence and the ratio of any two successive savings account balances will be the common ratio.

Then the common ratio is:

  $\begin{array}{l}r=\frac{\$22000}{\$20000},r=1.1\\ r=\frac{\$24,200}{\$22000},r=1.1\end{array}$

Therefore, the common ratio r is equal to 1.1.

ii.

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

The formula is $\$20000{\left(1.1\right)}^n$ .

Given information:

The value of r from part (a) is 1.1 and the given table is:

  

Explanation:

Consider the given table:

  

The value of r from part (a) is 1.1 it means every year, the balance becomes 1.1 times of the previous years’ balance.

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

  $a\cdot r^n=\$20000{\left(1.1\right)}^n$

Therefore, the formula is $\$20000{\left(1.1\right)}^n$ .

iii.

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

The answer is $\$370,623.341$ .

Given information:

The value of r from part (a) is 1.1 and the given table is:

  

Formula used: The sum of a finite geometric sequence $\left\{a_1,a_2,\cdots a_n\right\}$ with common ratio not equal to 1 is given by: $Sum=\frac{a_1\left(1-r^n\right)}{1-r}$

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 $Sum=\frac{a_1\left(1-r^n\right)}{1-r}$ ,

  $\begin{array}{c}Sum=\frac{\$20000\left(1-{\left(1.1\right)}^{11}\right)}{1-1.1}\\ =\$20000\left(18.53116\right)\\ =\$370,623.341\end{array}$

Therefore, the sum is $\$370,623.341$ .