Question

how do I do these summations again?
Find the sum of the first 12 terms of the sequence.

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

Answer 1

what is the difference between two consecutive terms?

Answer 2

n-5?

Answer 3

should be just -5

Answer 4

so 1 + -5 = -4
-4 + -5 = -9
-9 + -5 = -14 , etc

Answer 5

yes I got that part:) how do I set up the summation calculation for this though?

Answer 6

there is a formula for the sum of an arithmetic sequence

Answer 7

http://www.regentsprep.org/regents/math/algtrig/atp2/arithseq.htm

Answer 8

first you have to find the 12th term

Answer 9

we dont really need the 12th term do we? i mean its fine ... but theres a rewriting of it in terms of a1 that suffices

Answer 10

an = a1 + d(n-1)

a1 + a12 = a1 + a1+d(12-1)

Answer 11

to find a12 you can use the sequence formula
an = a1 + (n-1)*d

Answer 12

just a little trivia is all
\[\frac{n}{2}(2a_1+d(n-1))=\frac{n}{2}(a_1+a_n)\]

Answer 13

thats a simpler formula :)

Answer 14

algebrating it ...
\[n~a_1 + \frac{d}{2}(n-1)\]
tomato tomato lol

Answer 15

sorry I was in the bathroom:p
but I got n=2
a1=1
and d=-5
but what do I do with the summation sigma thing? I forgot how to set it up.
I added the first 12 numbers physically and I got -318

Answer 16

so we derived a formula for you
Sn = n/2 ( 2*a1 + d(n-1) )

Answer 17

S12 = 12/2 ( 2 * 1 + (-5)*(12-1) )

Answer 18

ahh ok that works to. is there a way to solve it using the sigma sign tho? \[\sum_{n=?}^{n}\]

Answer 19

if we know the rule that defines the sequence ..
\[\sum_{n=1}^{k} a_n\]
\[\sum_{n=1}^{k} d(n-1)+a_1\]
\[\sum_{n=1}^{k} dn-d+a_1\]
\[d\sum_{n=1}^{k} n+(a_1-d)\sum_{n=1}^{k}\]
and this requires still knowing how to sum up a sequence if consecutive integers to produce
\[d\frac{1+k}{2}+k(a_1-d)\]

Answer 20

so what exactly would I plug into the calculator?