# Math Essay

375 WordsJan 31, 20122 Pages
Here is how we think about it: It is \$100 for the first 10 feet. After that it is \$25 for each additional 10 feet. This tells us that it is arithmetic sequence because we are multiplying. N= the number of terms altogether N= 8 D= the common difference D=25 A1= the first term A1= 100 An= the second term An= a8 Next, we need to compute what a8 is. An= a8+ (n-1) d A8= 100+ (8-1)25 A8=100+7(25) A8=100+200 A8=300 Now that we know what A8 is, we need to know what the sum of the sequence is from A1 to A8. Sn= n (a1+a8) over2 S8= 8(100+300) over2 S8= 8(400) over2 S8= 4(400) = 1600 Thus, the cost to build a 90 foot CB tower is \$1600. A person deposited \$500 in a savings account that pays %5 annual interests that is compounded yearly. At the end of 10years, how much money will be in the savings account? Here is how we think about it: Each year 5% of the balance is added to the balance. If we let B= the balance, it would look like: B+ (.05) B B (1+.05) B (1.05) In other words, each year the existing balance is multiplied by 1.05. This repeated multiplication by the same number tells us we have a geometric sequence. First, we need to identify. N= the number of terms N= 10 R= the common ratio R= 1.05 A1= the first term A1= 500(1.05) =525 the balance at the end of 1 year In a savings account, the total balances at the end of each year form the sequence, so we don’t need to add up all the terms in the sequence. We just need to find out what the balance is at the end of 10 years, so we are look for the value of A10. An= a1 (rn-1) A10=525(1.05) square root of 9 A10= 525(1.55134) A10=814.46 Thus, the balance in the savings account at the end of 10 years will be 814.46. References:Mat126 Math in our World Bluman,1sted, Ashford assignment