The sum of the first n terms of a series is 1-(3/4). Obtain an expression for the nth term of the series. Prove that the series is geometric, and state the values of the first term and the common ratio. Please show workings

Answer 1

The n'th term is ar^(n-1)

Sn = a(1-r^n)/(1-r) = 1/4
4a(1-r^n) = 1-r
4a - 4ar^n = 1-r
4a - 4ar^(n-1)*r = 1-r
ar^(n-1) = (1-r-4a)/(4r)

I can't see how you can solve for r and a. There are many possible solutions.

n=1: a=1/4

n=2: a(1+r) = 1/4
a = 1, r = -3/4
a = -1/3, r = -7/4

n=3: a(1+r+r^2) = 1/4
...
I think there must be something missing here. Or a typo.