# Find the nth term of the series 1, 3 , 9 , 27...

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The first 4 terms of the series are 1, 3, 9 and 27. It is seen that `(3/1)= (9/3) = (27/9) = 3` . The ratio between consecutive terms is the same and equal to 3. This is the characteristic of a geometric series.

The nth term of a geometric series is `T_n = a*r^(n - 1)` where a is the first term and r is the common ratio.

**The nth term of the given series is 3^(n - 1)**

Looking at the series of numbers, you can see that any number is 3 times the number right before it. So with each additional term, you have multiplied a number by 3 one more time. This means that terms in the series can be rewritten as 3^0, 3^1, 3^2, 3^3, etc. So to find a relationship for the nth term will mean finding an exponent where the base is three. You'll notice that the first term has an exponent of 0, the second term has an exponent of 1, the third term has an exponent of 2, so the exponent is always one less than the number for the term. In other words (n-1) is the exponent for the nth term. So the required expression becomes 3^(n-1).

The given numbers 1, 3, 9, 27 belong to the geometric series as the ratio between the consecutive numbers is the same and eaual to 3.

The first term of the series is 1.

For a geometric series any term is given by a.r^(n-1) where a is the first term, r is common ratio and n is the term number.

* The nth term of the series is *1*3^(n-1) =

*3^(n-1)*