Exercises 35 and 37 in the “Real World Applications” Section on Page 280 of Mathematics in Our World.

Topics: Geometric progression, Sequence, Arithmetic progression Pages: 2 (432 words) Published: August 25, 2013
In this week’s assignment I will attempt complete exercises 35 and 37 in the “Real World Applications” section on page 280 of Mathematics in Our World. For each exercise, specify whether it involves an arithmetic sequence or a geometric sequence and use the proper formulas where applicable. I will try to format my math work as shown in the “week one assignment guide” provided to us and try to be concise in my reasoning.

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)
=...
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