Question

Given the arithmetic sequence an = 2 + 4(n - 1), what is the domain for n

Answer 1

usually you have positive or some times you can have positive and zero subscripts

Answer 2

do you have options ?

Answer 3

All integers
All integers where n ≥ 0
All integers where n > 1
All integers where n ≥ 1

Answer 4

I would tick the last option i think...

Answer 5

I honestly have no clue
the 2nd and 4th work for me :p

Answer 6

I seen some people use 2nd
I seen some people use the 4th

Answer 7

I agree with ganeshie8

Answer 8

http://www.regentsprep.org/regents/math/algtrig/atp2/arithseq.htm
according to this site a_1 is the first sequence

Answer 9

is the first in sequence*

Answer 10

for me all work haha
but i wil go with last option because of a silly reason \(a_n = a_1 + (n-1)d\) :
\(n=1\) gives \(a_1 = 2\) which is readily available in the nth term expression

Answer 11

i think this is a terrible question

Answer 12

because the other choices work work to

i mean I never seen negative subscripts but I suppose you could have them :p

Answer 13

completely agree !
n=0 disturbs a1 in the given pattern... thats a bit convincing to have the sequence start with index n=1.. but technically we can start the sequence anywhere we want yeah :)

Answer 14

so far all of these arithmetic sequence sites I looked at just now have started the sequence at a1 so @15Twilight I think @ganeshie8 would be correct

Answer 15

first term : \(a_0\)
second term : \(a_{-1}\)
third term : \(a_1\)
fourth term : \(a_{-2}\)

\(\ldots\)

nobody gona stop me from counting like this :P

Answer 16

when I say correct I mean I think you will get it right

Answer 17

ok thanks everyone!

Answer 18

I won't stop you @ganeshie8

Answer 19

:p

Answer 20

we can also rational numbers to count i guess hmm
first term : \(a_{1/1}\)
second term : \(a_{2/1}\)
third term : \(a_{1/2}\)
fourth term : \(a_{1/3}\)

\(\ldots\)

http://personal.maths.surrey.ac.uk/st/H.Bruin/image/rationals-countable.svg.png