Ellen C. Reed

Mat126 Problems 35, 37

Martha Stillman

July 19, 2013

Mat126, #35, 37 2

Mat126 Problem 35 and 37

I will be completing two math problems in this paper. Using the text book Mathematics in our World to provide me with both of my problems. I will be showing all of my work and explaining how I got to the answer. I will also determine if it is arithmetic sequence or geometric sequence.

Problem 35. A person hired a firm to build a radio tower. The firm charges $100.00 in labor for the first ten feet. Then the firm charges an additional $25.00 per 10 feet. That means the next 10 feet cost $125, then the next 10 feet will cost $ 150, and so on. How much will it cost to build a 90 foot tower? n= 9 an= a1 + (n-1) d d= 25 a9= 100 + (9-1) (25) a1= 100 =100 +8 (25) an= a9 (yet to be determined) =100 + 200 =300 we now know what a9 is, it is 300. The next step is the arithmetic sequence. =n (a1+a9)/2-1 =9(100+300) =9(400) =9(400) =3,600 The nth term $3,600 is what it will cost to build a 90-foot tower. In an arithmetic sequence you find the common difference.

Problem #37. A man deposited $500 in an annuity 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? n=10 =a1(r n-1) r=1.05 =525(1.05 11) =500(1.05) =525(1.7103389) =525 (balance of the first year) =