Solving Arithmetic Progressions

[nae99]

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

 

[Mark44]

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.

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.

 

[nae99]

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

 

[nae99]

That's not what it says. What you wrote is that the sum of the first 8 terms is 56.

ar^7 = 56

 

[Mark44]

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.

 

[Mark44]

That's not what it says. What you wrote is that the sum of the first 8 terms is 56.

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.

 

[nae99]

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

 

[nae99]

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

 

[PeterO]

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

 

[Mark44]

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.

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

 

[nae99]

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

 

[Bohrok]

...the 6th term is 4 times the sum of the 2nd and the 3rd.

a₆ = 4xa₂ + a₃

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?

 

[nae99]

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

 

[ArcanaNoir]

$$a_6=4(a_2+a_3)$$
Now can you write the sequence $a_1+a_2 ...=56$ without using the term $a_6$?

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.

 

[nae99]

$$a_6=4(a_2+a_3)$$
Now can you write the sequence $a_1+a_2 ...=56$ without using the term $a_6$?

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 $a_6$
this how i would write it:
$a_1+a_2+a_3+a_4+a_5+a_6+a_7+a_8=56$

 

[ArcanaNoir]

but a_6=? replace a_6 in the big sum equation by what stands where the ? is.

 

[nae99]

$$a_6=4(a_2+a_3)$$
Now can you write the sequence $a_1+a_2 ...=56$ without using the term $a_6$?

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.

 

[ArcanaNoir]

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?

 

[Mark44]

no, i do not know how to write it without using the term $a_6$
this how i would write it:
$a_1+a_2+a_3+a_4+a_5+a_6+a_7+a_8=56$

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?

 

[nae99]

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.

 

[ArcanaNoir]

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.

 

[Mark44]

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.

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.

 

[ArcanaNoir]

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!

 

[nae99]

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?

is it; d= a₁ + 7d

 

[Mark44]

is it; d= a₁ + 7d

No. How did you get that?

 

[nae99]

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.

 

[Mark44]

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

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:
$a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 = 56$

$a_6 = 4(a_2 + a_3)$

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.