# Arithmetic progression

1. Aug 11, 2011

### nae99

1. The problem statement, all variables and given/known data
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

2. Relevant equations

3. The attempt at a solution

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

2. Aug 11, 2011

### Staff: Mentor

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.

3. Aug 11, 2011

### nae99

a6 = 4xa2 + a3

4. Aug 11, 2011

### nae99

ar^7 = 56

5. Aug 11, 2011

### Staff: Mentor

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. Aug 11, 2011

### Staff: Mentor

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. Aug 12, 2011

### nae99

a6 = 4xa2xa3

8. Aug 12, 2011

### nae99

a1 + a2 + a3 + a4 + a5 + a6 + a7 + a8 = 56

9. Aug 12, 2011

### PeterO

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. Aug 12, 2011

### Staff: Mentor

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.

11. Aug 12, 2011

### nae99

oh ok
a6 = 4*a2*a3

12. Aug 12, 2011

### Bohrok

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. Aug 12, 2011

### nae99

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

i actually dont know what im doing

14. Aug 12, 2011

### 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.

15. Aug 12, 2011

### nae99

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

16. Aug 12, 2011

### ArcanaNoir

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

17. Aug 12, 2011

### nae99

if u said u used up tow sheets of paper and still cant solve it, how am i going to and i hardly know what to do.

18. Aug 12, 2011

### ArcanaNoir

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. Aug 12, 2011

### Staff: Mentor

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. Aug 12, 2011

### nae99

i did just the bacis maths before and now i'm doing a precalculus course.