# Given the sequence 3,9,27,...determine the nth term of the sequence and the 11th term .Given the sequence 3,9,27,...determine the nth term of the sequence and the 11th term .

*print*Print*list*Cite

The given sequence is:

3, 9, 27, ...

we notice that:

3 = 3^1,

9 = 3^2 and

27 = 3^3

Thus the sequence can be written as:

3^1, 3^2, 3^3, ...

Thus nth term of the sequence can be represented by general formula:

nth term = 3^n

Therefore:

11th term = 3^11 = 177147

Let's take the series, 3,9,27. This is not an AP as 27-9 is not equal to 9-3.

But the ratio of 27/9=9/3=3, therefore this is a geometric progression.

The nth term of a GP is ar^(n-1).

Here the first term is 3 therefore a=3 and as the common ratio is 3, r is also 3.

The nth term is 3*3^(n-1)

The 11th term would be 3*3^(11-1)=3*3^10=5904933*3=177147

3,8,27, .... is a sequence whose succesive terms are incresing by a common multiple of 3.

So the successive terms of the sequence are inreasing by a common ratio 3.

The 1st term of the GP is 3.

The nth term of the geometric sereis is an = 3*3^(n-1) = 3^n.

Therefore the 11 term is when n =11 or an =3^11

For the beginning, we'll prove that the sequence is a geometric progression. We'll form ratios from 2 consecutive terms of the given sequence:

9/3 = 3

27/9 = 3

.....................

We notice that all quotients are the same, so, the sequence is a geometric progression, whose first terms is a1 = 3 and the common ratio is r = 3.

We'll calculate a11:

a11 = a1*r^(11-1)

a11 = 3 * 3^10

a11 = 3^11

**a11 = 177147**

The standard formula for any term of a geometric progression is:

**an = a1*r^(n-1), where a1 is the first term of the progression and r is the common ratio.**