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doreent0722
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Need help on these questions I haven't been introduced yet. Please look at image I inserted in.
doreent0722 said:Need help on these questions I haven't been introduced yet. Please look at image I inserted in.
Jameson said:Hi doreent0722,
Welcome to MHB! :)
I don't see any images. Can you try typing one out or uploading them to a site like imgur.com?
Jameson said:Hi doreent0722,
Welcome to MHB! :)
I don't see any images. Can you try typing one out or uploading them to a site like imgur.com?
Jameson said:Ok, thank you for posting. First thing is, no need TO SHOUT :) We can read just fine.
What have you tried for the first question? We ask that you post one question at a time.
Is that $\displaystyle 2+\frac{5}{N}$ or $\displaystyle \frac{2+5}{7}$? I'm guessing the first one.
Let $N=1$ what do you get for the result?
doreent0722 said:Sorry for the caps its an OCD habit. I'm not yelling at you it just looks to me nicer.
I haven;t tried anything on those questions for yesterday was my first day in class for Algebra & trigonometry. Once again I'm so Sorry for the caps lettering. Will you please tutor on all of those questions so I can learn for my Monday class?
I'm sorry I don't know how to do this. Will you please show me how to by step by step to solve it.Jameson said:No worries about the caps. Now you know our custom. :)
I will try to help you figure it out but we don't give out answers so we're going to have to do this together.
You didn't comment on the two choices in my last post but let's assume that the sequence we have is:
$\displaystyle 2+\frac{5}{N}$
Ok let's start with $N=1$
$\displaystyle 2+\frac{5}{1}=2+5=7$. So the first term is 7.
If I want to find the third term, I plug in $N=3$.
$\displaystyle 2+\frac{5}{3}=\frac{6}{3}+\frac{5}{3}=\frac{11}{3}$
Using the same method, how would you get the 7th and 11th terms?
MarkFL said:I'll start at the other end and help with the third problem:
[box=blue]3). \(\displaystyle a_7 =\frac{1}{3}\), \(\displaystyle d= -\frac{2}{3}\), \(\displaystyle n=15\).
Find the sum of the arithmetic sequence $S_n$ that satisfies the stated conditions.[/box]
Okay, we will find helpful the following formula:
[box=red]Sum Of Arithmetic Progression
\(\displaystyle S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)\tag{1}\)[/box]
We are given $n$ and $d$, and we need to find $a_1$. Every time we reduce the index by 1 we increase the value of the term by $-d$. So, to get to $a_1$, we decrease the index by how much from $a_7$, and therefore, how much should be added to $a_7$ to get the value of $a_1$?
doreent0722 said:I'm sorry I don't know how to do this. Will you please show me how to by step by step to solve it.
doreent0722 said:Thanks so far but how do I solve for my answer?
MarkFL said:You need to find $a_1$...how many steps do you have to take back to get from $a_7$ to $a_1$?
doreent0722 said:Sorry this is my first homework from class yesterday. Instructor gave it to us to see where our weakness is and its also to be good on our behalf.
An arithmetic sequence is a sequence of numbers where the difference between consecutive terms is constant. This means that each term is obtained by adding (or subtracting) the same number, called the common difference, to the previous term.
To find the common difference of an arithmetic sequence, subtract any term from the term that comes after it. The result will be the common difference. For example, if the sequence is 2, 5, 8, 11, the common difference is 3 (5-2=3, 8-5=3, 11-8=3).
The formula for finding the nth term of 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.
To determine if a sequence is arithmetic, check if the difference between consecutive terms is constant. If it is, then the sequence is arithmetic. You can also use the formula an = a1 + (n-1)d to find the nth term and see if it follows the pattern of the sequence.
Arithmetic sequences can be used in various real-life situations such as calculating the growth of a population, determining the amount of money earned from investments with a fixed interest rate, and finding the distance traveled by an object moving at a constant speed. They can also be used in solving problems related to patterns and series in mathematics and computer programming.