Exercise 35:

A person hired to build a CB Radio tower. The firm charges $100 for labor for the first 10 feet. After that, the cost of the labor for each succeeding 10 feet is $25 more than the preceding 10 feet. That is, the next 10 feet will cost $125; the next 10 feet will cost $150, etc. How much will it cost to build a 90-Foot tower?

Formula & Breakdown:

This problem involves arithmetic sequence since labor cost for each successive 10 ft. remains constant at $25. The arithmetic sequence of the labor costs is 100, 125, 150, etc.

For finding the cost for building a 90 ft. tower, we sum the first 9 terms of the above sequence

TN=100+25•{N-1}

N=tens of feet

TN=cost to build that ten feet ex. to build 10 feet plug in 1 for N

T1=100+25 • {1-1}

T1=100

T2=100+25•{2-1}

T2=125

T3=100+25•{3-1}

T3=150

T4=100+25•{4-1}

T4=175

T5=100+25•{5-1}

T5=200

T6 = 100+25•{6-1}

T6=225

T7=100+25•{7-1}

T7=250

T8=100+25•{8-1}

T8=275

T9=100+25•{9-1}

T9=300

Then add all the TN’s together

100+125+150+175+200+225+250+275+300 = 1800 it will take $1800 to build the tower

Exercise 37:

A person deposited $500 in a saving account that pays 5% annual interest that is compounded yearly. At the end of 10 years, how much money will be in the savings account?

Formula used:

A = P (1 + r/n) ^ (nt)

= the initial deposit = the interest rate (percentage) = the years invested

Breakdown:

P = $500, t = 10 years, r = 5% or 0.05, n = 1 (annually compound) we need to find A (compounded amount)

A = 500(1 + 0.05/1) ^ (1*10)

=