arithmatic and geometric sequences

[joshm1975]

I am trying to find the answer to a question due tonight in my online course. I did simple addition to come up with the answer, but I need to understand how to get the answer with the correct formula. Please advise: A person hire3d a firm 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 preceeding 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?
I tried the formula an = a1+(n-1)d and the formula an = a1r n-1 but either I'm doing it wrong or I'm using the wrong formulas.

 

[mmm4444bot]

I tried the formula an = a1+(n-1)d

Did you use n = 9 ?

n is the number of 10-foot sections in 90 feet.

MY EDITS: changed symbol d to n

 

[joshm1975]

yes I used 9 for d. I also did simply addition to get the solution. with addition I came up with $1,800 but I cannot get that number using either formula. Am I not doing the formula correctly?

 

[Deleted member 4993]

I am trying to find the answer to a question due tonight in my online course. I did simple addition to come up with the answer, but I need to understand how to get the answer with the correct formula. Please advise: A person hire3d a firm 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 preceeding 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?
I tried the formula an = a1+(n-1)d and the formula an = a1r n-1 but either I'm doing it wrong or I'm using the wrong formulas.

You are trying to use the formula to find the n[sup:1wpv2a86]th[/sup:1wpv2a86] term (in your case, that would be cost of the just the last 10 ft. of the tower). As you have realized you need to use the summation formula. It should be in your text-book (You are following one - I hope).

If not - do a google search - and tell us which formulae you found. We will discuss from there....

 

[joshm1975]

the instructor wants me to tell if it is arithmatic sequence or geometric, but I'm not sure which it is. and yes I am following a text book and it does show how to do the sum of the first n terms of a geometric sequence, but I don't fully understand the concept. the formula for that is Sn = a1(1-rn) /1/r.

 

[mmm4444bot]

Whoops, I goofed. It's n that equals 9.

d = 25

a[1] = 100

a[n] = 100 + (n - 1)*25

The formula for adding a[1] through a[9] is:

S = n * (a[1] + a[n])/2

 

[mmm4444bot]

Okay. You haven't learned the definition of an arithmetic sequence or geometric sequence, yet.

When the elements in a sequence change by a fixed amount, then their sequence is arithmetic.

a = {100, 125, 150, 175, 200, … }

a is an arithmetic sequence because the elements change by a fixed amount (+25).

A geometric sequence is when elements change by a fixed multiplier (*r).

 

[joshm1975]

I thought it was an arithmatic sequence, but I still don't come up with 1800 using it. What are you coming up with?

 

[Deleted member 4993]

I thought it was an arithmatic sequence, but I still don't come up with 1800 using it. What are you coming up with?

You have 9 terms begining with 100 and the last term is 300

Then the sum is = 9 * (100 + 300)/2 = 1800

 

[mmm4444bot]

a[n] = 100 + (n - 1)*d

a[1] = 100 + (1 - 1)*25 = 100

a[9] = 100 + (9 - 1)*25 = 300

S = n * (a[1] + a[2])/2

S = 9 *(100 + 300)/2 = 1800

You need to start showing your work, if you want more help. I can't see what you're doing.

 

[BigGlenntheHeavy]

$\displaystyle Before \ you \ panic, \ take \ a \ deep \ breath \ and \ relax.$

$\displaystyle Now, \ assume \ you \ know \ nothing \ about \ progressions.$

$\displaystyle From \ the \ problem: \ Let \ a_1 \ = \ 100, \ then \ a_2 \ = \ 125, \ a_3 \ =150, \ etc.$

$\displaystyle or \ a_1 \ = \ 100+(0)(25)$
$\displaystyle \ \ a_2 \ = \ 100+(1)(25)$
$\displaystyle \ \ a_3 \ = \ 100+(2)(25)$
$\displaystyle \ \ a_n \ = \ 100+(n-1)(25)$

$\displaystyle a_n \ = \ a_1+(n-1)(25), \ a_1 \ = \ 100, \ n \ = \ number \ of \ terms.$

$\displaystyle Now, let \ 25 \ = \ d, \ d \ being \ the \ difference \ of \ two \ consecutive \ terms, \ d \ is \ positive$

$\displaystyle if \ the \ terms \ are \ increasing, \ negative \ if \ the \ terms \ are \ decreasing.$

$\displaystyle Hence, \ a_n \ = \ a_1+(n-1)d \ and \ we \ call \ this \ equation \ an \ arithmetic \ progression.$

$\displaystyle Now, \ we \ want \ to \ know \ how \ much \ it \ would \ cost \ to \ build \ a \ 90 \ ft. \ (n=9) \ tower.$

$\displaystyle Sum \ = \ a+(a+d)+(a+2d)+...+(l-2d)+(l-d)+l, \ a \ = \ first \ term, \ l \ = \ last \ term.$

$\displaystyle Sum \ = \ l+(l-d)+(l-2d)+...+(a+2d)+(a+d)+a, \ (sum \ is \ commutative).$

$\displaystyle Hence, adding, \ we \ get \ 2S \ = \ (a+l)+(a+l)+(a+l)+...+(a+l)+(a+l)+(a+l) \ = \ n(a+l)$

$\displaystyle Ergo, \ S \ = \ \frac{n}{2}(a+l) \ and \ last \ term \ (l) \ = \ a_1+(n-1)d, \ (n=9), \ (a=a_1).$

$\displaystyle Therefore, \ l \ = \ 100+(8)(25) \ = \ 300, \ \implies \ S \ = \ \frac{9}{2}(100+300) \ = \ \$1800.00$

$\displaystyle Note: \ We \ could \ have \ just \ added \ up \ the \ nine \ terms \ for \ the \ answer, \ but \ what \ if$

$\displaystyle we \ had \ 900 \ or \ 9000 \ terms?$

 

[joshm1975]

ok I'll show my work. this is my first week in this course and I havn't done math in 15 years, so I need to refresh on many things. I appreciate your help. While I check your work to see what I did wrong can you help me with one more question?
Question 37: A person deposited $500 in a savings 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?
I'm thinking this is a geometrical sequence because it has a multiplier of .05. My work so far is a1 = 500, r = .05, and n = 10.
Using the formula an = a1rn-1 I think it goes something like;
500 x .05 [10 - 1] = 500 x .05 [9] = 500 x .45 = 225. an = 22
However, something is not right because in this case the final product I came out with is 225.25 and I know that is too low. When I calculated it the slow way of multiplying .05 to each annual sum I came out with $855.17, which may not be exact but close. Can you tell me again what I'm doing wrong?
Thanks

 

[joshm1975]

I thought I posted my second question some time ago, but I guess not. I reviewed the answer to the previous question and I was wondering how you all came up with the 400 being divided by 2. Where did the 2 come from and how did you add an extra 100 to the 300?

 

[BigGlenntheHeavy]

$\displaystyle A \ = \ P\bigg(1+\frac{r}{n}\bigg)^{nt}$

$\displaystyle A \ = \ amount \ of \ accrual, \ P \ = \ principal \ (initial \ investment), \ r \ = \ interest \ rate, \ n \ = \ number$

$\displaystyle of \ times \ compounded \ yearly, \ and \ t \ = \ time \ in \ years.$

 

[joshm1975]

would that break down to a = 500 * 1 + .05 divided by 1 because that is the number of times the interest is compounded a year? Also, would the exponent nt translate into 1 x 10?

 

[BigGlenntheHeavy]

$\displaystyle A \ = \ 500(1.05)^{10}$

 

[BigGlenntheHeavy]

Where did the 2 come from?

$\displaystyle Are \ you \ capable \ of \ following \ a \ sequence \ of \ events? \ or \ must \ you \ be \ spoon-fed$

$\displaystyle like \ a \ 5 \ year \ old?$

 

[joshm1975]

I don't understand how your getting those numbers. can you elaborate in layman terms. This course did not give me an opportunity to refresh my algebra before throwing this at me. I am not behind yet, but I need to grasp this if I expect to keep up and pass. So yes, I do need to be spoon fed and find myself a tutor.

 

[BigGlenntheHeavy]

$\displaystyle Hey, \ if \ (2)(S) \ = \ (n)(a_1+l), \ then \ what \ does \ S \ equal?$

$\displaystyle Now, \ if \ you \ can't \ figure \ this \ out, \ you're \ over \ you \ head.$

 

[joshm1975]

I guess I'm over my head then because I would think I need to know the value of the letters