Solving Arithmetic Progressions

In summary: um...help you?if u said u used up tow sheets of paper and still can't solve it, how am i going to...um...help you?
  • #1
nae99
129
0

Homework Statement


The sum of the first 8 terms of an AP is 56, and the 6th term is 4 times the sum of the 2nd and the 3rd. Find the first term and the common difference


Homework Equations





The Attempt at a Solution



8th term = 56 6th term = 4x2nd+3rd
 
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  • #2
nae99 said:

Homework Statement


The sum of the first 8 terms of an AP is 56, and the 6th term is 4 times the sum of the 2nd and the 3rd. Find the first term and the common difference


Homework Equations





The Attempt at a Solution



8th term = 56
That's not what it says. What you wrote is that the sum of the first 8 terms is 56.
nae99 said:
6th term = 4x2nd+3rd
Instead of writing "6th term" and "2nd" and so on, use subscripts to indicate which term you're talking about. For example,
a1 = first term
a2 = second term

and so on.
 
  • #3
Mark44 said:
That's not what it says. What you wrote is that the sum of the first 8 terms is 56.

Instead of writing "6th term" and "2nd" and so on, use subscripts to indicate which term you're talking about. For example,
a1 = first term
a2 = second term

and so on.

a6 = 4xa2 + a3
 
  • #4
Mark44 said:
That's not what it says. What you wrote is that the sum of the first 8 terms is 56.



ar^7 = 56
 
  • #5
nae99 said:
a6 = 4xa2 + a3
That's still not right. It says "the 6th term is 4 times the sum of the 2nd and the 3rd"

You have 4 times the 2nd plus the 3rd.
 
  • #6
Mark44 said:
That's not what it says. What you wrote is that the sum of the first 8 terms is 56.

nae99 said:
ar^7 = 56
No. Write an expression for the sum of the first 8 terms.

For example, the sum of the first three terms would be a1 + a2 + a3. Since this is an arithmetic progression, you can write a2 and a3 in terms of a1.
 
  • #7
Mark44 said:
That's still not right. It says "the 6th term is 4 times the sum of the 2nd and the 3rd"

You have 4 times the 2nd plus the 3rd.

a6 = 4xa2xa3
 
  • #8
Mark44 said:
No. Write an expression for the sum of the first 8 terms.

For example, the sum of the first three terms would be a1 + a2 + a3. Since this is an arithmetic progression, you can write a2 and a3 in terms of a1.


a1 + a2 + a3 + a4 + a5 + a6 + a7 + a8 = 56
 
  • #9
nae99 said:
a6 = 4xa2xa3

Which of the four operation in arithmetic does "sum" represent? Also why is it NOT part of your attempt here?

* the four operations are addition, subtraction, multiplication and division
 
  • #10
nae99 said:
a6 = 4xa2xa3
In addition to what PeterO says below, you should not be writing 'x' to indicate multiplication, because x is used so often as the name of a variable.

PeterO said:
Which of the four operation in arithmetic does "sum" represent? Also why is it NOT part of your attempt here?

* the four operations are addition, subtraction, multiplication and division
 
  • #11
Mark44 said:
In addition to what PeterO says below, you should not be writing 'x' to indicate multiplication, because x is used so often as the name of a variable.

oh ok
a6 = 4*a2*a3
 
  • #12
nae99 said:
...the 6th term is 4 times the sum of the 2nd and the 3rd.

nae99 said:
a6 = 4xa2 + a3

nae99 said:
a6 = 4xa2xa3

You're close; how would you write out "the sum of the 2nd and the 3rd"? Then, how would you write out 4 times that sum?
 
  • #13
Bohrok said:
You're close; how would you write out "the sum of the 2nd and the 3rd"? Then, how would you write out 4 times that sum?

a6 = 4*a1 + a2 * a1 + a2 + a3

i actually don't know what I am doing
 
  • #14
[tex]a_6=4(a_2+a_3)[/tex]
Now can you write the sequence [itex] a_1+a_2 ...=56[/itex] without using the term [itex]a_6[/itex]?

I spent a good two sheets of paper trying to solve this and I never got it to work out. I'm eager to see what to do with it.
 
  • #15
ArcanaNoir said:
[tex]a_6=4(a_2+a_3)[/tex]
Now can you write the sequence [itex] a_1+a_2 ...=56[/itex] without using the term [itex]a_6[/itex]?

I spent a good two sheets of paper trying to solve this and I never got it to work out. I'm eager to see what to do with it.

no, i do not know how to write it without using the term [itex]a_6[/itex]
this how i would write it:
[itex] a_1+a_2+a_3+a_4+a_5+a_6+a_7+a_8=56[/itex]
 
  • #16
but a_6=? replace a_6 in the big sum equation by what stands where the ? is.
 
  • #17
ArcanaNoir said:
[tex]a_6=4(a_2+a_3)[/tex]
Now can you write the sequence [itex] a_1+a_2 ...=56[/itex] without using the term [itex]a_6[/itex]?

I spent a good two sheets of paper trying to solve this and I never got it to work out. I'm eager to see what to do with it.

if u said u used up tow sheets of paper and still can't solve it, how am i going to and i hardly know what to do.
 
  • #18
nae99 said:
if u said u used up tow sheets of paper and still can't solve it, how am i going to and i hardly know what to do.

Don't worry, there's a whole team of people here to help! :) Just because I'm not getting it doesn't mean it's too hard, it just means I have more to learn on this matter, just as you do. By the way, what math class are you taking now, and what have you taken before?
 
  • #19
nae99 said:
no, i do not know how to write it without using the term [itex]a_6[/itex]
this how i would write it:
[itex] a_1+a_2+a_3+a_4+a_5+a_6+a_7+a_8=56[/itex]

Now, since this is an arithmetic difference, each term can be found by adding the same amount to the preceding term. Let's say that the common difference is d. Then a2 = a1 + d, and Then a3 = a2 + d = (a1 + d) + d = a1 + 2d.

Can you figure out how to write the sum of the 8 terms that you wrote so that it involves only a1?
 
  • #20
ArcanaNoir said:
Don't worry, there's a whole team of people here to help! :) Just because I'm not getting it doesn't mean it's too hard, it just means I have more to learn on this matter, just as you do. By the way, what math class are you taking now, and what have you taken before?

i did just the bacis maths before and now I'm doing a precalculus course.
 
  • #21
Mark44, not to clutter up the thread, but, have you solved this all the way out? I tried what you're suggesting and some other things but I never got the solution. I'm not trying to say you're wrong, I'm just dying to know if someone has solved it.
 
  • #22
ArcanaNoir said:
I spent a good two sheets of paper trying to solve this and I never got it to work out. I'm eager to see what to do with it.

ArcanaNoir said:
Mark44, not to clutter up the thread, but, have you solved this all the way out? I tried what you're suggesting and some other things but I never got the solution. I'm not trying to say you're wrong, I'm just dying to know if someone has solved it.
I hadn't done so before, but I have now, and have checked my answer. It turns out that the first term of the sequence is negative, and the common difference is positive.
 
  • #23
OMG I see what I did. I had the answer all along but I forgot when I checked a_2+a_3 to multiply it by 4 to compare it to a_6. Doh doh doh!
 
  • #24
Mark44 said:
Now, since this is an arithmetic difference, each term can be found by adding the same amount to the preceding term. Let's say that the common difference is d. Then a2 = a1 + d, and Then a3 = a2 + d = (a1 + d) + d = a1 + 2d.

Can you figure out how to write the sum of the 8 terms that you wrote so that it involves only a1?

:uhh::confused:
is it; d= a1 + 7d
 
  • #25
nae99 said:
:uhh::confused:
is it; d= a1 + 7d
No. How did you get that?
 
  • #26
Mark44 said:
No. How did you get that?

ok i gave up because i have no idea what i am doing...i don't know how to do it.
 
  • #27
nae99 said:

Homework Statement


The sum of the first 8 terms of an AP is 56, and the 6th term is 4 times the sum of the 2nd and the 3rd. Find the first term and the common difference

The Attempt at a Solution



8th term = 56 6th term = 4x2nd+3rd

nae99 said:
ok i gave up because i have no idea what i am doing...i don't know how to do it.
You have these two equations:
[itex]a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 = 56[/itex]

[itex]a_6 = 4(a_2 + a_3)[/itex]

If you know how an arithmetic progression is defined, you should be able to write both equations in terms of a1 and the common difference d. If you can do that, you will have two equations in two unknowns, a1 and d. Solve the system of equations for these variables.
 

1. What is an arithmetic progression?

An arithmetic progression is a sequence of numbers where the difference between each pair of consecutive terms is constant. This difference is called the common difference, denoted by d.

2. How do you find the nth term in an arithmetic progression?

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

3. Can an arithmetic progression have a negative common difference?

Yes, an arithmetic progression can have a negative common difference. This means that the terms in the sequence are decreasing.

4. What is the sum of an arithmetic progression?

The sum of an arithmetic progression can be found using the formula: Sn = (n/2)(a1 + an), where Sn represents the sum, n is the number of terms, a1 is the first term, and an is the nth term.

5. What are some real-life examples of arithmetic progressions?

Arithmetic progressions can be seen in various situations, such as the increase in salary over the years, the growth of a population, or the depreciation of a car's value each year.

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