Question

What is the sum of a 12-term arithmetic sequence where the last term is 13 and the common difference is -10?

Answer 1

Here \(n = 12\).

\(l = 13\)
and \(d = -10).

Firstly find out first term by using:
\[l = a + (n-1)d\]

Answer 2

* \(d = -10\)

Answer 3

\[l = a + (n-1)d\]

Answer 4

So.. 13 = a +(12-1)-10
13 = a+(11)-10
13= a+1
-1
a=12
???

Answer 5

\[S_n=\frac{ n }{ 2 }[2a+(n-1)d]\]

Answer 6

use the formula waterineyes gave you to find a,
since you were already given the values of n and d in the question...
\[l=a+(n-1)d\]
\[13=a +(12-1) \times -10\]
so find a and put into the formula...

Answer 7

So, 13=12+(12-1)x-10
13= 12+(11)x-10
13=23x-10
+10
23=23x
x=1
O.o Now, I'm superr lost lol

Answer 8

\(\bf d=-10\) --> Common difference

\(\bf a_{13}= a + 12d = 13\)

Notice, by writing term 13 in terms of 'a' and 'd', we can now solve for 'a' by plugging in -10 for d.
\[\bf a+12d=13 \rightarrow a+12(-10)=13 \implies a = 133\]So now we have the first term, 'a', which means we can now use the summation formula to find the sum of the first 13 terms:\[\bf S _{n}=\frac{ n }{ 2 }[2a+(n-1)d]\]Plug in 133 = a, -10 = d, and 13 = n, and evaluate.

Can you do that?

@Imckeex3

Answer 9

I'm doing something wrong..

sn = 13/2[2(133)+(13-1)-10]
13/2[266+12-10]
13/2(268)...

Answer 10

it is supposed to be 12/2 not 13/2......

Answer 11

\[\bf S_{13} = \frac{ 13 }{ 2 }[2(133)+(13-1)(-10)]=\frac{ 13 }{ 2 }[266+(-120)]=\frac{ 13 }{ 2 }[146]=949\]
That's what I got.

@lmckeex3

Answer 12

S(n)=Sum of the n terms in the sequence
n=number of terms in the sequence = 12
a=the 1st term in the sequence
d=common difference

Answer 13

Oh sorry I did 13, I forgot that they only wanted the sum of the first 12 lol.

Answer 14

Just change 13 to 12 and you should get an answer of 936.

Answer 15

The choices are wrong.

Answer 16

-_-

Answer 17

there's a mistake....a is supposed to be 123 not 133....

Answer 18

omg lol

Answer 19

if you make a the subject of the formula i gave above using the formula waterineyes gave,
you get a to be 123...
\[13=a + (12-1) \times -10\]

Answer 20

if you correct the mistake, what do you get now??

Answer 21

by x.. do you mean *?

Answer 22

yes i mean multiplication....

Answer 23

ddnt we say that 13 was actually 12?
I'm assuming by x you mean *?
so 12=123+(12-1)x-10
12 = 123+11-10
12= 134 * -10
12 = -1340

I'm just going to guess lol. I've completed the quiz. This is the last question.

Answer 24

\[S _{12} = \frac{ 12 }{ 2 }[2(123) + (12-1) \times -10]\]

Answer 25

do you see it now??

Answer 26

6[(246)+(11)(-10)

816

I wasn't using pemdas correctly. THANK YOU!!!!

Answer 27

you are welcome....(:

Answer 28

You have found a wrong there..

Sorry, I am late..

\[13 = a + (11)(-10) \implies a = 13 + 110 = a = 123\]

Answer 29

Here when you find \(a\) and \(l\) here, why you go for that long formula??

Just use:

\[Sum = \frac{n}{2}(a + l)\]

Answer 30

^^ That would have been helpful an hour ago lol.

Answer 31

\[S_{12} = \frac{12}{2}(123 + 13) = ??\]

Answer 32

I think this would have been helpful 13 hours ago, when you got offline.. :P

Answer 33

lol. :x I fell asleep. Thanks to everyone for their help tho. I got 100.