Find the nth tern of the sequence 1,1,1,1

The nth term is Tn = (-1) (n-1)

Explanation:

sequence = 1, - 1, 1, - 1, 1, - 1, 1, - 1, ...?

This is a periodic, oscillating sequence. It is also something called a Geometric Progression where each term is a ratio. In your description you were expecting an arithmetic progression it seems - but it is not so.

For n=1

First term = 1

For n=2

Second term = - 1

For n=3

Third term = 1

For n=4

Fourth term = - 1

In every successive term the power is one less than the number of the particular term itself (for n=1th term the power was 0)

So we have this sequence:

(-1) 0, (-1) 1, (-1) 2, (-1) 3, (-1) 4, (-1) 5, ...1,-1,1,-1,1,-1, ...

To express this as a summation:

∑n=1∞ (-1) (n-1)

So the nth term of the sequence in question will be (-1) (n-1) where n is the particular term of the sequence.

There is also a very simple formula you can use to arrive at the nth term of a geometric progression:

Tₙ = a⋅ (r (n-1))

n is the nth term of the sequence

r is the common ratio

a is the first term of the sequence

So inserting these values:

Tn = 1⋅ ((-1) (n-1))

Tn = (-1) (n-1)