Question

I want to find the nth term of a sequence.
n=1,2,3,4,5
terms=81,27,9,3,1
a(n)=

Answer 1

You're good with sequences, right?

Answer 2

id divide by 3 each time meself ....

Answer 3

They are powers of 3, but what is the sequence?

Answer 4

an = a1*r^(n-1) ; n>0

Answer 5

It looks like the a_1 term is 81 though.

Answer 6

yes 81 * r^1-1 = 81

Answer 7

Yes, sorry by not mentioning this. but I know the answer should be a(n)(1/3)^n-1, but on the back of the book it says 243/3^n, I'm wondering where the 243 came from

Answer 8

if you allow n to start at zero you have to start at a0 instead of a1; is 81*3 243?

Answer 9

81*3 = 243?

Answer 10

well it says n starts at 1, and yes 81*3=243

Answer 11

both ways are equal

\[a_n=a_1r^{n-1}\text{ same as }a_n=a_0r^n\]

Answer 12

all youve done is moved your index back by 1

Answer 13

so since it is 81,27,9,3,1 and its 81*3=243 so a(n)=243/3^n
is it true that if 81,20.25,5.0625,1.265625,.316406 (times 1/4)
so is it 81*4=324 so a(n)=324/4^n

Answer 14

yes

Answer 15

but, there isnt a wrong one and a right one ... its just a format

Answer 16

@amistre64 : what does r in the formula signify?

Answer 17

Common Ratio

Answer 18

@thushananth01 r=Common Ratio, forexample in this questions r=1/3

Answer 19

Would these be correct?\[\left\{ n \right\}_{n =1 }^{\infty} = 1, 2 , 3, 4, 5, ...\]\[\left\{ 243 ( \frac {1}{3} )^n \right\}_{n = 1}^{\infty} = 81, 27, 9, 3, 1, ...\]

Answer 20

looks fine to me

Answer 21

The nth term for the first sequence is simply n. The nth term for the second sequence is 243/(3^n).

Answer 22

r is a general term for the common ratio