Question

find the geometric series with r=3 and s=364

Answer 1

\[s_n = \frac{a_1(r^{n} -1)}{r-1}\]
\[364 = \frac{a_1(3^{n} -1)}{2}\]
need more info to find a_1 and n

Answer 2

thats all they give u

Answer 3

you may do trial and error then

Answer 4

1 + 3 + 3^2 + ....

Answer 5

oh the question says find a1 for the geometric series

Answer 6

you need a reference point, at least one number in the sequence

Answer 7

see when the sum reaches 364

Answer 8

how so

Answer 9

since there is not enough info, this is not a direct q... they may want you test few sums and figure out the series

Answer 10

if psble, take a screenshot of the q u seeing and attach... it helps to confirm :)

Answer 11

mmkaay

Answer 12

Now that makes sense. Did you notice, you missed the 6 in \(S_6\)

Answer 13

lol ohhh no i didnt sorryy

Answer 14

its okay :)

Answer 15

So, you're given r=3, and n = 6.

Answer 16

plug them in the formula given by dumbcow and solve \(a_1\)

Answer 17

okay

Answer 18

\(\large s_n = \frac{a_1(r^{n} -1)}{r-1}\)

\(\large s_6 = \frac{a_1(3^{6} -1)}{3-1}\)

\(\large 364 = \frac{a_1(3^{6} -1)}{2}\)

Answer 19

you can solve \(a_1\)

Answer 20

would it be 243?

Answer 21

it wont be that big for sure

Answer 22

try again, use ur calculator

Answer 23

oh you are given n ... i was going to say there are multiple answers
given an "n" you can find a "a_1" that works

Answer 24

idk i got 364

Answer 25

the a1 is just there?

Answer 26

\(\large 364 = \frac{a_1(3^{6} -1)}{2} \)

Answer 27

multiply 2 both sides

Answer 28

\(\large 364 \times 2 = a_1(3^{6} -1) \)

Answer 29

\(\large \frac{364 \times 2}{(3^6-1)} = a_1\)

Answer 30

wolfram says it equals 1

http://www.wolframalpha.com/input/?i=%28364*2%29%2F%283%5E6-1%29