Yes, you are correct. :-)
Hello, cityismine!
I suspect the typo is in the statement of the problem . . .
(a) Formula: .a(n) .= .a(1)·r^{n-1}The first term of a geometric sequence is 27, and the sixth term is 1024/9
(a) Find the seventh term.
(b) Find the sum of the first five terms of this sequence.
We have: .a(1) = 27, . a(6) = 1024/9
Then: .a(6) = a(1)·r^5 . → . 1024/9 = 27r^5 . → . r^5 = 1024/243
. . Hence: .r .= .4/3
Therefore: .a(7) .= .a(6)·r .= .(1024/9)(4/3) .= .4096/27
. . . . . . . . . . . . . . . . . . . r^n - 1
(b) Formula: .S(n) .= .a(1)·---------
. . . . . . . . . . . . . . . . . . . . r - 1
. . . . . . . . . . . . . . .(4/3)^5 - 1 . . . 781
Hence: . S(5) .= .27·-------------- .= .-----
. . . . . . . . . . . . . . . . 4/3 - 1 . . . . . .3
Were those the answers they expected?