Question

Find the sum of the first 12 terms of the sequence. Show all work for full credit.

1, -4, -9, -14, . . .

Answer 1

I would start by finding the 12th term, right?

Answer 2

Well, find the rest of the terms.

Answer 3

Do you see the pattern?

Answer 4

It's easy if you find the common difference in the pattern, and realize it will most likely be a arithmetic progression.

Answer 5

Remember for common difference, \[t_2-t_1 = t_3-t_2\] where t represents term and the subscript is term in the progression

Answer 6

It's -5, right?

Answer 7

That's the pattern

Answer 8

Right on :), but there is a short cut, to figuring out the 12th term, or 100th, or 5000th, or millionth, you get the point.

Answer 9

hehe, yeah :)

Answer 10

Yep, we're subtracting 5.

Answer 11

So four our problem here, we will use \[t_n = a+(n-1) \times d\] a is the first term in the sequence, d is the common difference and n is the term you're looking for.

Answer 12

It's pretty neat, we can derive it from your problem, but do this for me, use this and than compare it by doing it the tedious way :P

Answer 13

so tn = 1 + (n-1) * -5

Answer 14

oops forgot 12
tn= 1 + (12 - 1) * -5

Answer 15

Yeah :)

Answer 16

so -60?

Answer 17

Mhm, nope.

Answer 18

Try again

Answer 19

oh my goodness - 54

Answer 20

got my order of operations wrong :P

Answer 21

Yeah :)

Answer 22

Is that the whole answer? I feel like I am missing something.. hm

Answer 23

Now you can even try by doing it without the little formula we just used

Answer 24

And you'll get the same answer ;)

Answer 25

That's it, your 12th term is -54

Answer 26

Oh, I see it's asking you to list the first 12 sums, you can do that as well, just -5, from each term, but you also have this neat formula now.

Answer 27

So if you realized, the main thing here was noticing the common difference, and it's a breeze.

Answer 28

@iambatman