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)

if you explain the the question more correctly i might be able to answer this better

step-by-step explanation:

the solution is x=1

step-by-step explanation:

we are given our   equation as

let's assume

left side function is

right side function is

now, we can draw graph

and then we can get intersection point

the intersection point will be solution

so, the solution is

x=1