**The Geometric Sequence**

Numbers are said to be in Geometric Sequence if there is a common ratio between any two consecutive terms.

**Example: **

In the sequence of the following numbers:

2, 4, 8, 16, 32, .....

The ratio between any two consecutive numbers is 2, i.e. 4/2 is same as 8/4.

In the above example of terms in geometric sequence, the common ratio is 2.

The **common ratio** between any two consecutive terms is denoted as “r”.

Denote first term as “a”.

Then the terms in a geometric sequence can be denoted as

**a, ar, ar ^{2}, ar^{3}, .............ar^{n – 1}**

Recall that numbers are in Geometric Sequence if there is a common ratio between any two consecutive terms.

The common ratio is denoted by “r”

Let the terms of the geometric sequence be denoted as

**a, ar, ar ^{2}, ar^{3}, ............ , ar^{n – 1}**

Then, the nth term or general term of a G.S. is

**a r ^{n – 1} **

**Note: **

**The power in ‘r’ is one less than the nth term of a geometric sequence. **

**Therefore, 2 ^{nd} term is ar, 3^{rd} term is ar^{2}, 4^{th} term is ar^{3}, **

**Example: **

**1. The first term of a geometric sequence is 1/3. The common ratio is 3. Find the six terms of the G.S.**

**Solution: **

First term, a = 1/3, common ratio, r = 3. Now, the other terms are:

Second term is ar = (1/3) × 3 = 1,

Third term is ar2 = (1/3) × 3^{2} = 3,

Fourth term is ar3 = (1/3) × 3^{3} = 9,

Fifth term is ar4 = (1/3) × 3^{4} = 3^{3} = 27,

Sixth term is ar5 = (1/3) × 3^{5} = 3^{4} = 81

**2. The fifth term of a geometric sequence is 625. The first term is unity. Find the 3rd term of the sequence.**

**Solution: **

Recall that in a geometric sequence, the fifth term is denoted as ar4.

Now,

ar^{4} = 625, and a, the first term is 1, i.e. a = 1.

1 × r^{4} = 625, r^{4} = 625, r^{4} = 5^{4}, r = 4.

So, the third term denoted as ar^{2} is

1× 4^{2} = 16

Consider a geometric sequence with a common ratio ‘r’ and first term ‘a” in which the n terms are

a, ar, ar^{2}, ar^{3}, ....... ar^{n-1}

Let Sn denote the sum of the n terms of the above geometric sequence

Then,

S_{n} = a (r^{n} – 1)/(r – 1), if r > 1, and

S_{n} = a (1 - r^{n})/ (1 - r), if r < 1

**Example: **

**1. What is the sum of the first 10 terms of a geometric sequence in which the 5 terms are 2, 4, 8, 16, and 32?**

**Solution: **

From the terms **2, 4, 8, 16, and 32, **it can be seen that the

First term a = 2, and the common ratio, r = 2.

Since r > 1, apply the first of the above formulas to find sum of the 10 terms of the given geometric sequence

S_{10} = 2 (2^{10} – 1)/ (2 – 1) = 2 (2^{10} – 1)

Since, 2^{10} is not a very large number, we will write its expansion:

2^{10} = 1024

So, S_{10} = 2 (1024 – 1) = 2 × 1023 = 2046

**2. What is the sum of the first 10 terms of a geometric sequence in which the first 3 terms are 1/3, 1/9 and 1/27?**

**Solution: **

We have to find the sum given below:

1/3 + 1/9 + 1/27 +........... + 1/310

First term is a = 1/3, second term is 1/9,

Therefore common ratio is r = 1/9/1/3 = 1/3

Since r < 1 (as r = 1/3 < 1),

From the above formulas, apply the second one to find the sum of the first 10 terms of the given geometric sequence

S_{n} = a (1 - r^{n})/ (1 - r), if r < 1

S_{10} = [1/3 (1 – 1/3^{10})]/ [(1 – 1/3)]

[1/3 (3^{10} – 1)]/ [3^{10} × 2/3] = (3^{10} – 1)/2 × 3^{10}

Therefore, 1/3 + 1/9 + 1/27 +........... + 1/310 = (3^{10} – 1)/2 × 3^{10}