Question

can someone explain the geometric series formula? please

Answer 1

S = a( R^n- 1) /R-1

a -> first term.
R -> a_{n+1} / a_{n}
S-> Sum
n _> number of terms

a_{n} = aR^n

the term that n can be found using this formula.

Answer 2

The geometric series is often written as:\[\sum_{n=0}^{\infty}t^n\]

Answer 3

so if give 4−12+36−.... what goes where in the formula

Answer 4

More specifically, if the first term (a) is 4, how do we get from that 4 to the next term, -12?

Similarly, if the 2nd term is -12, how do we obtain the 3rd term, 36?

We're talking about r, the "common ratio"

Answer 5

yea what im trying to figure out is how to get the common ratio so i can determine the sum of the 6th term

Answer 6

Find the sum of the first 666 terms in this geometric series:
4 - 12 + 36 - ... the sum would be -728 correct?

Answer 7

6th term not 666 term

Answer 8

Common ratio : R = a_{n+1} / a_{n}

Answer 9

-12/4 = -3

-12 * -3 = 36 (verified)

Answer 10

You wanted to know how to find the common ratio.

If the first term (a) is 4, how do we get from that 4 to the next term, -12?

Similarly, if the 2nd term is -12, how do we obtain the 3rd term, 36?

We're talking about r, the "common ratio"

The first and second terms are 4 and -12, respectively.

The common ratio is -12/4 = ??

We can check this by doing a similar operation on the 2nd and 3rd terms:

The second and third terms are -12 and 36. Dividing 36 by -12, we get what?

This result is your common ratio, r.