Question

Which is the 23rd term of the arithmetic series 5, 14, 23, …

Answer 1

do you know the formula? :)

Answer 2

i know that it goes up by 9

Answer 3

5 + (23 * 9) = 212 is the answer

Answer 4

but she needs to know how to do it, just just the answer.

Answer 5

cant be doesnt match up to the answers given

Answer 6

yea kreshnik is right

Answer 7

What are the choices?

Answer 8

247

203

269

181

Answer 9

i even counted them out and the closetest i got was 248

Answer 10

\[\LARGE a_n=a_1+(n-1)d\]
\[\LARGE d=a_n-a_{n-1 }\]
so... we have:
d=14-5=9
d=23-14=9
so d=9

and a_1=5
now we substitute:
\[\LARGE a_{23}=\underbrace{5}_{a_1}+(23-1)\underbrace{9}_{d}\]
so we have:
\[\LARGE a_{23}=5+198\]
can you do it now ? :)

Answer 11

203

Answer 12

I was about to say that haha

Answer 13

lol ok. thanks.

Answer 14

what if it gives me numbers other than a sequence like 25th term of the arithmetic sequence where a1 = 8 and a9 = 80

Answer 15

@Kreshnik you genius!

Answer 16

is it the same equation

Answer 17

ahh.. Google crashed :) I hate this so much. You have no Idea.

Here we write it again.

Answer 18

what u mean write it again

Answer 19

so we have:
a1=8 and a9=80
we have to find a25
but we first have to find difference "d"
how we find it?
Easy !
let's solve for a9
\[\LARGE a_9=a_1+(9-1)d\]
\[\LARGE 80=8+8d\]
\[\LARGE 80-8=8d\]
\[\LARGE 8d=72\]
\[\LARGE d=\frac{72}{8}\]
\[\LARGE d=9\]
now we have difference d=9
we solve it for a25
\[\LARGE a_{25}=a_1+(25-1)d\]
\[\LARGE a_{25}=8+(25-1)9\]
and there you go :)

Answer 20

what is it gives me the first and last term and asks for a term

Answer 21

well sometimes you have a question like:
a_3=20 a_8=50
find a_12
do you know how to do it ? :P