# Math Help - Geometric Series

1. ## Geometric Series

Determine whether or not the indicated geometric series converges. If so, find the value to which it converges.

t1 = 81 and t5 = 1 (Two answers)

2. Hello, Trentt!

Determine if this geometric series converges.
If so, find the value to which it converges.

t
1 = 81 and t5 = 1 (Two answers)

The first term is: .t
1 .= .a .= .81 .[1]

The fifth term is: .t
5 .= .ar^4 .= .1 .[2]

. . . . . . . . . . . . . ar^4 . . . . 1
Divide [2] by [1]: .------ .= . ---- . . . . r^4 .= .1/81
. . . . . . . . . . . . . . a . . . . . 81

. . Hence: .r .= .±1/3

In both cases, |r| < 1 . . . Therefore, both series converge.

If r = 1/3, the sum is: .S .= .81/(1 - 1/3) .= .243/2

If r = -1/3, the sum is: .S .= .81/(1 + 1/3) .= .243/4

3. I have no idea what you just did. I don't understand how you get the ratio with just two terms

4. Originally Posted by Trentt
I have no idea what you just did. I don't understand how you get the ratio with just two terms
You divide two sides,

For example,

x^2 = 7y^2
x=y

Divide left side with left side, divide right side with right side.And keep = sign in middle,

Thus,

Code:
x^2      7y^2
----  =  -------
x             y
And you get,
x=7y

5. Nope, that didn't clear it up.

6. Originally Posted by Trentt
Nope, that didn't clear it up.
Here you do this one.

xy^2 = xy^2
And,
x=y

Divide both sides like I did and see what you get.
Can you do this problem?

7. Nope.

8. Originally Posted by Trentt
Nope.
Did you try doing the problem. Post what you know how to do.

9. In the prior problems, before

t1 = 81 and t5 = 1

they gave us t1 and r, which I understood how to do. For this problem, I assume you have to find r, since the formula is t1/1-r
I don't know how to find r.

10. Originally Posted by Trentt
In the prior problems, before

t1 = 81 and t5 = 1

they gave us t1 and r, which I understood how to do. For this problem, I assume you have to find r, since the formula is t1/1-r
I don't know how to find r.
ok, so i don't see what's so hard to understand about Soroban's solution, but let's baby step through it.

in a geometric series, we have all terms of the form ar^(n-1) where a is the first term, r is the common ratio and n = 1,2,3,4,5....

so the first term is ar^(1-1) = ar^0 = a
the second term is ar^(2-1) = ar^1 = ar
the third term is ar^(3-1) = ar^2
the fourth term is ar^(4-1) = ar^3
the fifth term is ar^(5-1) = ar^4
and so on...

so from above, we can write the first term as a, so we have t1 = a, and we can write the fifth term as ar^4.

so t1 = 81 and t5 = 1 so t5/t1 = 1/81, but also t1 = a and t5 = ar^4, so we have:

.t5....=....1....=....ar^4
----......----.......------
.t1.........81..........a

so we get 1/81 = ar^4/a and we solve for r
on the right, the a's cancel, so we are left with 1/81 = r^4
so we get r = fourthroot(1/81) = +/- 1/3

with me so far?

11. anyway, now you have two values for r, so you can substitute them into that formula you mentioned to find the two answers. but Soroban already did that for you

12. YES. Thank you.