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.