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.

DENOTATION OF TERMS IN A GEOMETRIC SEQUENCE

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, ar2, ar3, ………….arn – 1

HOW TO FIND THE NTH TERM OR GENERAL TERM OF A GEOMETRIC SEQUENCE?

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, ar2, ar3, ………… , arn – 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, 2nd term is ar, 3rd term is ar2, 4th term is ar3,

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) × 32 = 3,

Fourth term is ar3 = (1/3) × 33 = 9,

Fifth term is ar4 = (1/3) × 34 = 33 = 27,

Sixth term is ar5 = (1/3) × 35 = 34 = 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, ar4 = 625, and a, the first term is 1, i.e. a = 1. 1 × r4 = 625, r4 = 625, r4 = 54, r = 4. So, the third term denoted as ar2 is 1× 42 = 16

HOW TO FIND THE SUM OF N TERMS OF A GEOMETRIC SEQUENCE?

Consider a geometric sequence with a common ratio ‘r’ and first term ‘a” in which the n terms are a, ar, ar2, ar3, ……. arn-1. Let Sn denote the sum of the n terms of the above geometric sequence. Then, Sn = a (rn – 1)/(r – 1), if r > 1, and Sn = a (1 – rn)/ (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 S10 = 2 (210 – 1)/ (2 – 1) = 2 (210 – 1). Since, 210 is not a very large number, we will write its expansion: 210 = 1024. So, S10 = 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 Sn = a (1 – rn)/ (1 – r), if r < 1, S10 = [1/3 (1 – 1/310)]/ [(1 – 1/3)], [1/3 (310 – 1)]/ [310 × 2/3] = (310 – 1)/2 × 310. Therefore, 1/3 + 1/9 + 1/27 +……….. + 1/310 = (310 – 1)/2 × 310.