Question

Question about finding sum of the arithmetic sequence? So i know the equation is Sn=a(1-r^n)÷(1-r), but usually the common ratio is multiplying by something. Is the formula different when the common ratio is adding?

Answer 1

It's called a common *difference* when arithmetic sequences are involved, and there's a formula for that, too ^_^

Answer 2

Ohhh okay so i would use a different formula then? thank you!

Answer 3

Yes, you know what it is?

Answer 4

uhmmm I do not ;o I was going to try googling it but i'm not sure what it would be called? finding the sum of a arithmetic sequence with a common difference? >_<

Answer 5

When an arithmetic sequence is concerned, the sum is given by this formula instead:
\[\Large S_n = a_1 + (n-1)d\]where \(a_1\) is the first term
\(n\) is the number of terms and
\(d\) is the common difference

Answer 6

okay! i can take it from here, thank you again! ^_^

Answer 7

WAIT NO!

Answer 8

Sorry, my mistake, that's just the nth term.
\[\Large a_n = a_1 + (n-1)d\]

Answer 9

o-o

Answer 10

To get the sum, the formula is

Answer 11

\[\Large S_n = \frac{n(a_1+a_n)}{2}\]

Answer 12

Where n is the number of terms, a1 is the first term, an is the last term,

Answer 13

hmm well my problem is What is the sum of the arithmetic sequence 3, 9, 15..., if there are 36 terms? so would 15 be the last term? or would i first find the term of 36?

Answer 14

No...the last term is an, find it with this formula\[\Large a_n = a_1 + (n-1)d\]

Answer 15

okay, thank you again!

Answer 16

Don't thank me until it's over D:

Answer 17

What's the 36'th term?

Answer 18

\[\Large a_n = a_1 + (n-1)d\]
in this formula, take n = 36

Answer 19

uhm i got 213 as the 36th term? ;o

Answer 20

oh okay, good ^_^
\[\large a_{36}=213\]
Now find the sum
\[\Large S_n = \frac{n(a_1+a_{n})}{2}\] again, n = 36

Answer 21

I got 3,888 :3

Answer 22

Brilliant.

Well done ^_^

Answer 23

Thank you! ^_^ thank you again for helping me x3

Answer 24

No problem.