# Arithmetic sequences

1. Jun 26, 2006

### kippy

i need help- arithmetic sequences

There many arithmetic sequences which seventh term equals 5. prove all of them have the same sum of their first 13 elemnets. find the sum

i found the sum was 65 but i dont know how to prove it.

Last edited: Jun 26, 2006
2. Jun 26, 2006

### AKG

Let tn = a + d(n-1) define an arithmetic sequence. Then t7 = a + d(7-1) = a + 6d = 5. Let sn represent the sum of the first n terms. What is sn in terms of a, d, and n? What is sn when n = 13?

3. Jun 26, 2006

### 0rthodontist

Write the expression for the 6th term as a function of the 7th term and the common difference d. What do you get if you add the 6th term to the 8th term? What do you get if you add the 5th term to the 9th term?

4. Jun 26, 2006

### d_leet

Well an arithmetics sequence is going to start with some initial element call it a0 and have some constant call it d that is added to each element in order to find the next element, so if our initial element is a0 then the general term an will be defined by

an = a0 + n*d

Note that the way I have defined this sequence the 7th term will actually be a6.

So in your case you have that a6 = 5

Now a simpler way to look at an arithmetic sequence is that if you are given an term of the sequence then to find the next or previous element all you need to do is simply add or subtract d (that is the difference between terms) respectively.

So since you have the 7th term equal to 5 then the 8th would be 5+d, the 9th 5+2d and so on, and for the previous elements you would have the 6th being 5-d and so on. Can you see how you can prove that the sum is always 65 now?

5. Jun 26, 2006

### kippy

6. Jun 26, 2006

### kippy

(5-d)+(5+d)

(5-2d)+(5+2d)

7. Jun 26, 2006

### d_leet

No I said nothing about ratios between ds in different sequences and I gave no reasons why the sum is 65, I merely tried to lay the foundation necessary to prove that the sum is always 65. think about it this way

1st term...
.
.
.
4th term = 5 - 3d
5th term = 5 - d - d = 5 -2d
6th term = 5 - d
7th term = 5
8th term = 5 + d
9th term = 5 + d + d = 5 + 2d
10th term = 5 + 3d
.
.
.
13th term...

That is what your arithmetic sequence will look like, so to prove that the sum is 65 all you really need to do is add each of the terms together starting with term 1 and ending with term 13. The to this problem is really to notice what the terms on either side of a given term will look like and what happens when you compute their sum. Does this help you any more?

8. Jun 26, 2006

### kippy

yes it does

9. Jun 27, 2006

### arunbg

For mathematical rigour (though unimaginative) you can also try,
$$a_7=a+6d\;\;\;\; S_{13} =\frac{13}{2}(2a+12d)=13\times a_7$$
ie, the sum of first 13 terms depends only on the 7th term .

Last edited: Jun 27, 2006
10. Jun 27, 2006

### 0rthodontist

It is perfectly rigorous to advance a verbal argument on the fact that the sum of each pair of terms n away from the 7th term must be 10.

11. Jun 27, 2006

### arunbg

Yes, but that's imaginative rigour :)