Question

how do you find a^n for the following geometric sequence
a^1=6,r=-2

Answer 1

Howz the sequence goes ?

Answer 2

a(n+1) = a0 * -2n

i think

Answer 3

hard to tell where to start; some suggest a0 some suggest a1

Answer 4

yeah,the question doesn't gives all information.

Answer 5

it gives all the pertenant info, but the method seems to be more material specific

Answer 6

if its starts at a1 then we have:

an = 6*-2(n-1)

a1 = 6*-2(0) = 6

Answer 7

if it starts at a0 then:

an = -3*-2n

Answer 8

I am really having trouble understanding the problem,what is \[a^n\] means here? Is the sequence like \[ a,a^2,a^3,a^4 \cdots \] ?

Answer 9

i think its a typo for a_n

Answer 10

\[a_n=a_1*-2(n-1)\]

Answer 11

But when \[ a_1=1 \] (first term) and r=common ratio r=-2,shouldn't \[a_n = a_1 \times r^{(n-1)}=6 \times (-2)^{(n-1)} \] ...

Answer 12

and how is this \[ a_n=a_1*-2(n-1) \] holds? What exactly I am missing here ?:/

Answer 13

geometric, if i recall is a multiplier between terms; the functional notation results in an exponential result tho

Answer 14

if we are just going from term to term tho, its a multiplier;

Answer 15

Sorry but I simply can't understand your argument .. I am talking about http://en.wikipedia.org/wiki/Geometric_progression and you ? :)

Answer 16

"Such a geometric sequence also follows the recursive relation"

Answer 17

i might be messing it up tho, but that recursive relation is what i was considering

Answer 18

yep,and that relation is

\[ a_n =r \times a_{(n-1)} \]

but how does that explain
\[ a_n=a_1*-2(n-1) \] ????!!!

Answer 19

\[a_1 = 6\]
\[a_2 = 6(-2)=-12\]
\[a_3 = 6(-2)(-2)=24\]
\[a_3 = 6(-2)(-2)(-2)=-48\]
\[a_n = a_1*(-2)^{n-1}\]

i see what i did, i forgot to put it as an exponent

Answer 20

Aha,thats makes sense now :)

Answer 21

im old and feeble, thats my excuse lol

Answer 22

Time to change your diapers I guess ?:P

Answer 23

still got some load to them, maybe in an hour ;)

Answer 24

lolz :D Cheers ! :)