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17 March, 04:17

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

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  1. 17 March, 04:42
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    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)
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