Question

Find S 32 for the series 5+8+11...I haven't done arithmetic sequence in a while...please don't just give me an answer...I want to say it's (32/2)*(5+31*3)???

Answer 1

Difference between terms is 3. Initial value is 5. So, Nth term will be 5+(N-1)*3.

Answer 2

You can find 32nd term by using that expression. Or, for that matter, any term.

Answer 3

Are you looking for the sum of all terms up to the 32nd term?

Answer 4

YES :-)

Answer 5

PLEASE :-)

Answer 6

It is given by n(1st term + nth term)/2. So, you kind of have it right. but, need to re-look at your nth term addition.

Answer 7

32*(5 + 5 + 31*3)/2 will be the right value.

Answer 8

okay, something is bothering me though but i guess I'll quickly revise why the formula is true thank you for all your help!!!

Answer 9

What is bothering you? :)

Answer 10

the formula is saying when you add up all the values up to the nth term you simply add n times the addition of the first & last term and half it...is it b/c the first & last term yield+(2nd+2nd to last term) + etc yields twice as much so you half it?

Answer 11

I GOT IT THANKS!!!

Answer 12

Well....it is like this. You can write the sum of terms as:
5 + 5 + 3*1 + 5 + 3*2 + 5 + 3*3 + ......

Taking all the 5's out, you get 5*n. Then, you are left with:
1 + 3*1 + 3*2 + 3*3 + ...... + 3(n-1).

That can be simplified further to get to the "formula".

This link has the proof:
http://www.purplemath.com/modules/series6.htm

Answer 13

5+8+11+...+92+95+98=(5+98)+(8+95)+(11+92)+...=added the first term& the last term, N times but we half it b/c we only have (n/2) pairs=n*( a1+an )/2

Answer 14

Yeah.....it relies on breaking down the series into simplier summations that are then proved to equate that way.