Which is the 9th term where do i stop adding?

[aisha]

find the sum of the first 9 terms in the series

9+11+13+15+17+19+...

is the 9 t0 or t1?
what is the 9th term?
...21+23+25+27

25 or 27??

 

[chroot]

The first nine terms are... the first nine terms.

9
11
13
15
17
19
21
23
25

Count them.

- Warren

 

[dextercioby]

And you have two ways to add them:brute force (simply add the 9 #-s) or elegantly,using the formulas for an arithmetic progression.

Daniel.

 

[aisha]

ok I got that part thanks, the next part says the sum of the first 5 terms can be found by S2=[n(t1+t5)]/2

create a formula to find the sum of the first 8 terms I just replace t5 in the above equation with t8 is this correct?

The next part says create a formula to find the sum of the first n terms how do u do this?

 

[dextercioby]

Is that "n" the step...?I think so.There's a nice and very old proof for the general formula.Do you know it (has it been taught to you at school...?)?

Daniel.

 

[dextercioby]

Nope,i checked.That "n" is not the step,but the # of terms.Sorry.

Daniel.

P.S.So in your example that "n" should be first 5,then 8,then arbitrary "n".

 

[Jameson]

The common formula for calculating arithmetic series should be in your book, or you should have discussed it? Have you ever seen a general formula?

 

[aisha]

yes i know n is the number of terms

A formula to find the sum of the first 8 terms would be

S2=[n(t1+t8)]/2 where n=8 and t1 and t8 are taken from the series in my first post .

How do u create a formula to find the sum of the first n terms?
Yes I've seen the one in my book there are two sn= (n/2)(a+tn) when d is unknown
and sn=(n/2)[2a+(n-1)d] when tn is unknown? Which one do I put for this question because I think from the previous part that shows how to find the sum of 5 terms and my formula for finding the sum of 8 terms needs to be used to derive sum of first n terms...

 

[dextercioby]

There's the simple proof.U can define an arithmetical progression through the recurrence relation:
a_{1}+a_{n}=a_{2}+a_{n-1}

Use that and the trick
S=a_{1}+a_{2}+...+a_{n-1}+a_{n}
S=a_{n}+a_{n-1}+...+a_{2}+a_{1} And simply add:
-----------------------------------------------
2S=(a_{1}+a_{n})+(a_{2}+a_{n-1})+...+(a_{n-1}+a_{2})+(a_{n}+a_{1})

Use the recurrence relation and a simple counting to write
2S=n(a_{1}+a_{n})
S=\frac{n}{2}(a_{1}+a_{n})

Simple,huh...?

Daniel.

 

[aisha]

Yes thanks I got it!