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)

Printable View

- April 5th 2007, 06:37 PMTrenttGeometric 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) - April 5th 2007, 07:46 PMSoroban
Hello, Trentt!

Quote:

Determine if this geometric series converges.

If so, find the value to which it converges.

t1 = 81 and t5 = 1 (Two answers)

The first term is: .t1 .= .a .= .81 .**[1]**

The fifth term is: .t5 .= .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

- April 5th 2007, 08:07 PMTrentt
I have no idea what you just did. I don't understand how you get the ratio with just two terms

- April 5th 2007, 08:25 PMThePerfectHacker
- April 5th 2007, 08:27 PMTrentt
Nope, that didn't clear it up.

- April 5th 2007, 08:40 PMThePerfectHacker
- April 5th 2007, 09:03 PMTrentt
Nope.

- April 5th 2007, 09:07 PMThePerfectHacker
- April 5th 2007, 09:13 PMTrentt
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.

- April 5th 2007, 09:46 PMJhevon
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? - April 5th 2007, 09:52 PMJhevon
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

- April 6th 2007, 06:41 AMTrentt
YES. Thank you.