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

In summary, when Gamma counted the money he deposited each month, he found that there are possible values for money deposited that are 184/m where m = 1,2,3,4,………………….
  • #1
Gamma
357
11
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.
 
Last edited:
Physics news on Phys.org
  • #2
All three look correct, though you may want to formulate #3 in terms of all factors of 18,400 (cents).
 
  • #3
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?
 
  • #4
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
 
  • #5
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.
 
  • #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??
 
  • #7
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.
 
  • #8
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
 

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

What is an arithmetic sequence?

An arithmetic sequence is a sequence of numbers where the difference between any two consecutive terms is constant. This constant difference is called the common difference and is denoted by d.

What is the formula for finding the nth term in an arithmetic sequence?

The formula for finding the nth term in an arithmetic sequence is:
an = a1 + (n-1)d
where an is the nth term, a1 is the first term, and d is the common difference.

How do you find the common difference in an arithmetic sequence?

To find the common difference in an arithmetic sequence, subtract any two consecutive terms. The result will be the common difference, denoted by d.

What is the 35th term in the arithmetic sequence 2, 5, 8, 11, ...?

To find the 35th term in this sequence, we can use the formula:
a35 = 2 + (35-1)3 = 2 + 102 = 104
Therefore, the 35th term in this sequence is 104.

What is the sum of the first 57 terms in the arithmetic sequence 2, 5, 8, 11, ...?

The sum of the first 57 terms in this sequence can be found using the formula:
S57 = (57/2)(2 + 57(3)) = 57(59) = 3363
Therefore, the sum of the first 57 terms in this sequence is 3363.

Similar threads

Replies
1
Views
2K
Replies
8
Views
1K
3
Replies
93
Views
12K
3
Replies
93
Views
8K
2
Replies
51
Views
8K
2
Replies
61
Views
9K
2
Replies
43
Views
11K
3
Replies
77
Views
14K
2
Replies
46
Views
12K
2
Replies
46
Views
6K
Back
Top