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)
Hello, Trentt!
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
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?