I disagree that the sum is 6.
common ratio of the first two terms ...
common ratio of the 2nd and 3rd terms ...
infinite sum ...
Q.: The numbers 1/ t, 1/ (t - 1), 1/ (t + 2) are the first, second and third terms of a geometric sequence.
Find (i) the value of t, (ii) the sum to infinity of the series.
Eq.: S¥ = a/ 1 - r
Ans.: (i) t = 1/ 4, (ii) S¥ = 6
Attempt:
(i) U2/ U1 = U3/ U2
(1/ (t - 1))/ (1/ t) = (1/ (t + 2))/ (1/ (t - 1))
(1/ (t - 1)) (1/ (t - 1)) = (1/ t) (1/ (t + 2))
(1/ (t^2 - 2t + 1)) = (1/ (t^2 + 2t))
(1/ (t^2 - 2t + 1 - t^2 - 2t))
1/ (-4t + 1) … this is where I lose the flow, slightly. Can someone help me confirm if my next move is permissible?
1/ (4t + 1)
1 = 4t
t = 1/ 4
(ii) S¥ = a/ 1 - r
a = 1/ (1/ 4) = 4, i.e. 1/ t
r = U2/ U1 or U3/ U2 = (1/ ((1/ 4) - 1))/ 4 = (1/ (-3/ 4))/ 4 = (-4/ 3)/ 4 =
(-4/ 3)(1/ 4) = -4/ 12 = -1/ 3
4/ (1 - (-1/ 3)) … This is where I think I may have gone wrong.
4/ (1 + (1/ 3)) … I think I need to have 1 - (1/ 3) here but I can’t see how to bring that about.
4/ (4/ 3)
4 (3/ 4)
12/ 4 = 3 The answer here should be 6 but I can only get 3.
If I can get 1 - (1/ 3), above, I’ll have (4/ (2/ 3)), and that gives 6 but as I have shown, the question does not work out that way for me. Have I missed something? Thank you.