Chapter 8.1, Problem 127E

(a).

To determine

To compute: the first eight terms of the sequence.

Answer to Problem 127E

First eight terms of the sequence are as follows

  $\underset{\_}{5050,5100.5,5151.5,5203.02,5255.05,5307.60,5360.68,5414.28}$ .

Explanation of Solution

Given information:

The sequence representing balance in the account after $n$ quarters when initial deposited amount is $5000 and rate of interest is 4% compounded quarterly is as follows

  $A_n=5000{\left(1+\frac{0.04}{4}\right)}^n$

Calculation:

Need to calculate first 8 term of following sequence.

  $A_n=5000{\left(1+\frac{0.04}{4}\right)}^n$

Calculate first term by substituting $n=1$ .

  $A_1=5000{\left(1+\frac{0.04}{4}\right)}^1$

  $=5000{\left(1+0.01\right)}^1$

  $=5000\times 1.01$

  $=5050$

Calculate second term by substituting $n=2$ .

  $A_2=5000{\left(1+\frac{0.04}{4}\right)}^2$

  $=5000{\left(1+0.01\right)}^2$

  $=5000\times {\left(1.01\right)}^2$

  $=5100.5$

Calculate third term by substituting $n=3$ .

  $A_3=5000{\left(1+\frac{0.04}{4}\right)}^3$

  $=5000{\left(1+0.01\right)}^3$

  $=5000\times {\left(1.01\right)}^3$

  $=5151.5$

Calculate fourth term by substituting $n=4$ .

  $A_4=5000{\left(1+\frac{0.04}{4}\right)}^4$

  $=5000{\left(1+0.01\right)}^4$

  $=5000\times {\left(1.01\right)}^4$

  $=5203.02$

Calculate fifth term by substituting $n=5$ .

  $A_5=5000{\left(1+\frac{0.04}{4}\right)}^5$

  $=5000{\left(1+0.01\right)}^5$

  $=5000\times {\left(1.01\right)}^5$

  $=5255.05$

Calculate sixth term by substituting $n=6$ .

  $A_6=5000{\left(1+\frac{0.04}{4}\right)}^6$

  $=5000{\left(1+0.01\right)}^6$

  $=5000\times {\left(1.01\right)}^6$

  $=5307.60$

Calculate seventh term by substituting $n=7$ .

  $A_7=5000{\left(1+\frac{0.04}{4}\right)}^7$

  $=5000{\left(1+0.01\right)}^7$

  $=5000\times {\left(1.01\right)}^7$

  $=5360.68$

Calculate eighth term by substituting $n=8$ .

  $A_8=5000{\left(1+\frac{0.04}{4}\right)}^8$

  $=5000{\left(1+0.01\right)}^8$

  $=5000\times {\left(1.01\right)}^8$

  $=5414.28$

Hence first eight terms of the sequence are as follows

  $\underset{\_}{5050,5100.5,5151.5,5203.02,5255.05,5307.60,5360.68,5414.28}$ .

(b).

To determine

To calculate: the balance in account after 10 years.

Answer to Problem 127E

After 10 years balance in the account will be $\underset{\_}{\$7444.32}$ .

Explanation of Solution

Given information:

The sequence representing balance in the account after $n$ quarters when initial deposited amount is $5000 and rate of interest is 4% compounded quarterly is as follows

  $A_n=5000{\left(1+\frac{0.04}{4}\right)}^n$

Calculation:

Expression of account balance is as follows

  $A_n=5000{\left(1+\frac{0.04}{4}\right)}^n$

Need to calculate balance after 10 years

As $n$ represents number of quarters , so for 10 years , number of quarters is $10\times 4=40$

Calculate fortieth term by substituting $n=40$ .

  $A_{40}=5000{\left(1+\frac{0.04}{4}\right)}^{40}$

  $=5000{\left(1.01\right)}^{40}$

  $=7444.32$

Hence after 40 quarters that is after 10 years balance in the account will be $\underset{\_}{\$7444.32}$ .