Chapter 9.4, Problem 37E

Solution

i.

To state: The first term and the common ratio r if Mr. O’Hara deposits $120 at the end of each month into an account that pays 7 percent interest compounded monthly. After 10 years, the balance in the account, in dollars, is

  $120{\left(1+\frac{0.07}{12}\right)}^0+120{\left(1+\frac{0.07}{12}\right)}^1+\cdots +120{\left(1+\frac{0.07}{12}\right)}^{119}$

The first term 120 and common ratio $r=1+\frac{0.07}{12}$ .

Given information:

After 10 years, the balance in the account, in dollars, is

  $120{\left(1+\frac{0.07}{12}\right)}^0+120{\left(1+\frac{0.07}{12}\right)}^1+\cdots +120{\left(1+\frac{0.07}{12}\right)}^{119}$

Explanation:

The given sequence is:

  $120{\left(1+\frac{0.07}{12}\right)}^0+120{\left(1+\frac{0.07}{12}\right)}^1+\cdots +120{\left(1+\frac{0.07}{12}\right)}^{119}=a_1+a_2+\cdots +a_n$

The balance shows a geometric sequence. The first term of the sequence is:

  $\begin{array}{c}{a}_1=120{\left(1+\frac{0.07}{12}\right)}^0\\ =120\left(1\right)\\ =120\end{array}$

Then the common ratio is:

  $\begin{array}{c}r=\frac{a_2}{a_1}\\ =\frac{120{\left(1+\frac{0.07}{12}\right)}^1}{120{\left(1+\frac{0.07}{12}\right)}^0}\\ =1+\frac{0.07}{12}\end{array}$

Therefore, the common ratio r is equal to $r=1+\frac{0.07}{12}$ .

ii.

To calculate: The sum of geometric sequence of the balance.

  $120{\left(1+\frac{0.07}{12}\right)}^0+120{\left(1+\frac{0.07}{12}\right)}^1+\cdots +120{\left(1+\frac{0.07}{12}\right)}^{119}$

The answer is $\$20,770.18$ .

Given information:

After 10 years, the balance in the account, in dollars, is

  $120{\left(1+\frac{0.07}{12}\right)}^0+120{\left(1+\frac{0.07}{12}\right)}^1+\cdots +120{\left(1+\frac{0.07}{12}\right)}^{119}$

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 sequence:

  $120{\left(1+\frac{0.07}{12}\right)}^0+120{\left(1+\frac{0.07}{12}\right)}^1+\cdots +120{\left(1+\frac{0.07}{12}\right)}^{119}$

The sequence starts from 0 to 119, so there are 120 terms in the sequence.

Substitute the values in the sum formula $Sum=\frac{a_1\left(1-r^n\right)}{1-r}$ ,

  $\begin{array}{c}Sum=\frac{120\left(1-{\left(1+\frac{0.07}{12}\right)}^{120}\right)}{\left(1-\left(1+\frac{0.07}{12}\right)\right)}\\ =\$20,770.176\end{array}$

Therefore, the sum is $\$20,770.18$ .