Arithmetic Sequences: Examples & Solutions for 35th, 57th Terms

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Homework Help Overview

The discussion revolves around arithmetic sequences, specifically focusing on finding terms within these sequences and addressing related numerical problems. Participants explore examples of arithmetic sequences, the calculation of specific terms, and the interpretation of conditions related to divisibility.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants present various examples of arithmetic sequences to meet specific term requirements, questioning the validity of their approaches. There is also exploration of conditions for divisibility and remainder calculations, with some participants suggesting alternative interpretations of the problems.

Discussion Status

The discussion is active, with participants providing different methods and interpretations for the arithmetic sequence problems. Some guidance has been offered regarding the formulation of the problems, but no consensus has been reached on the interpretations of the divisibility conditions.

Contextual Notes

Participants are navigating through potential ambiguities in the problem statements, particularly regarding the conditions for divisibility and the definitions of terms in arithmetic sequences. There is an ongoing examination of assumptions related to the problems presented.

Gamma
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1. Give an example of an arithmetic sequence such that the 35th term is 4,207?

I used the general form of an arithmetic seq. an = a1 + (n-1)d and found that,

a1 = 25, and d = 123

Does this look ok? I had to use some trial and error since we have two unknowns.2. What is the 57th smallest whole number that has a remainder of 2 when it is divisible by 4 and 6.

I am getting 686. 3. Mark saved money in his bank. The first month he put 11$ in the bank. Every month thereafter he deposited more money, and it was the same amount each month. When he counted the money he totaled 195.00$ .What are possible amounts of money he could have deposited each month?

Solution:
Lets assume that he continued to deposit for m number of months after the first month. Also let's assume that he deposited x dollars each month.

I am feeling something is missing in this problem. Anyway this is how I did this.

195 = 11 + m * x

m * x = 184
x = 184/m where m is a whole number.

The possible values of money deposited are 184/m where m = 1,2,3,4,………………….

Some possible values for the money deposited are: $184, $92, $61.3, $46, ………..
Thanks a lot,

Gamma.
 
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All three look correct, though you may want to formulate #3 in terms of all factors of 18,400 (cents).
 
1. Give an example of an arithmetic sequence such that the 35th term is 4,207?]

A way to do this without "trial and error" is to note that dividing 4207 by 35-1= 34 gives 123 with remainder 25: 4207= 25+ 34(123).

2. What is the 57th smallest whole number that has a remainder of 2 when it is divisible by 4 and 6.
I think you mean "when it is divided by 4 and 6". 12 is the smallest number that is )= 24 and so the first "smallest whole number that has a remainder of 2 when it is divided by 4 and 6" is 26. In general the nth such number is 24n+ 2. Taking n= 57, 24(57)+ 2= 1370. Am I misinterpreting?
 
For the first problem, you can as well take the first term as 4173 and the common difference as 1.
i.e.
a= 4173
d = 1
The arithmetic sequence in this case would be 4173,4174,4175...4207.
spacetime
Physics
 
HallsofIvy said:
I think you mean "when it is divided by 4 and 6". 12 is the smallest number that is )= 24 and so the first "smallest whole number that has a remainder of 2 when it is divided by 4 and 6" is 26. In general the nth such number is 24n+ 2. Taking n= 57, 24(57)+ 2= 1370. Am I misinterpreting?

Isn't 2 the smallest whole number that has a remainder of 2 when it is divided by 4 or 6.
 
I think you mean "when it is divided by 4 and 6". 12 is the smallest number that is )= 24 and so the first "smallest whole number that has a remainder of 2 when it is divided by 4 and 6" is 26. In general the nth such number is 24n+ 2. Taking n= 57, 24(57)+ 2= 1370. Am I misinterpreting?

Words can confuse. I thought 12 is the smallest whole number which satisefies our condition and 12 is therefore the first such whole number.

Here are the numbers in ascending order

12, 24, 36, ......an

an = 12n, for n= 57, a57 = 684

57 th such number is 684+2 = 686. By the way this tutorial is on arith. series and so this solution make sense i guess.:smile:

HallsofIvy: I don't understand why you did what you did. Even if 24 is the smallest such number, then you should be getting an = 12(n+1)
a57 = 696 and so the answer would have been 696+2= 698.


Isn't 2 the smallest whole number that has a remainder of 2 when it is divided by 4 or 6.

Can there be a remainder when a smaller number is devided by a larger number?? 2/4 = 1/2??
 
Gamma said:
Can there be a remainder when a smaller number is devided by a larger number?? 2/4 = 1/2??
Why not? Let's think about it:
2 = 0 x 4 + 2. So when 2 is divided by 4, 0 is the quotient, and 2 is the remainder... :wink:
------------------
So in general, if n < p then when divinding n by p, you get 0 as the quotient, and n as the remainder.
 
2 = 0 x 4 + 2


I am sure the man who wrote this problem did not think about it.


While we are at this can I ask another question regarding sequences?

What is meant by index and value of a sequence? Is index n and value an?

This is the question: Using the index of a sequence as the domain and the value of the sequence as the range, is a sequence a function?

I think the answer is yes.

Thanks

Gamma
 

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