Geometric Mean
Ms. G. Martin
The terms between any two non-consecutive terms of a
geometric sequence are called geometric means.
If A1, A2, …..An-1, An is a geometric sequence, then the numbers
A2,…..,An-1 are the geometric means between A1 and An.
Example 1. Insert three geometric means between 1 and
81
Steps Solution
1. Identify the given information
(1, ___, ___, ___, 81)
Here, A1 = 1 and An (which is A5) =
81
2. Find the common ratio using
the formula An = A1rn-1
.
A1 = 1 and A5 = 81, we have
An = A1rn – 1
81 = (1)r4
81 = r4
4
81 = r
3 = r
3. Solve for the three geometric
means using the common ratio.
1, ___, ___, ___, 81
A1 = 1
A2 = 1(3) = 3
A3 = 3(3) = 9
A4 = 9(3) = 27
So, the three geometric means between 1 and 81 are 3, 9, and 27.
Example 2. Insert five geometric means between 8 and 512.
Steps Solution
1. Identify the given information. A1 = 8 and A7 = 512
2. Solve the common ratio using the
formula
An = A1rn-1.
A1 = 8 and A1= 512, we have
An = A1rn-1
512 = 8(r)7 – 1
512 = 8r6
64 = r6
6
64 = r
2 = r
3. Solve for the five geometric means
using the common ratio.
8, A2, A3, A4, A5, A6, 512
A1 = 8
A2 = 8(2) = 16
A3 = 16(2) = 32
A4 = 32(2) = 64
A5 = 64(2) = 128
A6 = 128(2)= 256
A7 = 512
So, the five geometric means between 8 and 512 are 16, 32, 64, 128, 256.
Example 3. Find the geometric mean between 5 and 8.
Steps Solution
1. Identify the given information Hence m = 5 and n = 8
2. Solve the geometric mean using
the formula of 𝑚𝑛 .
𝑚𝑛 = 5. 8 = 40
= 4. 10
= 2 10
So the geometric mean between 5 and 8 is 2 10.

Geometric mean

  • 1.
  • 2.
    The terms betweenany two non-consecutive terms of a geometric sequence are called geometric means. If A1, A2, …..An-1, An is a geometric sequence, then the numbers A2,…..,An-1 are the geometric means between A1 and An.
  • 3.
    Example 1. Insertthree geometric means between 1 and 81 Steps Solution 1. Identify the given information (1, ___, ___, ___, 81) Here, A1 = 1 and An (which is A5) = 81 2. Find the common ratio using the formula An = A1rn-1 . A1 = 1 and A5 = 81, we have An = A1rn – 1 81 = (1)r4 81 = r4 4 81 = r 3 = r
  • 4.
    3. Solve forthe three geometric means using the common ratio. 1, ___, ___, ___, 81 A1 = 1 A2 = 1(3) = 3 A3 = 3(3) = 9 A4 = 9(3) = 27 So, the three geometric means between 1 and 81 are 3, 9, and 27.
  • 5.
    Example 2. Insertfive geometric means between 8 and 512. Steps Solution 1. Identify the given information. A1 = 8 and A7 = 512 2. Solve the common ratio using the formula An = A1rn-1. A1 = 8 and A1= 512, we have An = A1rn-1 512 = 8(r)7 – 1 512 = 8r6 64 = r6 6 64 = r 2 = r
  • 6.
    3. Solve forthe five geometric means using the common ratio. 8, A2, A3, A4, A5, A6, 512 A1 = 8 A2 = 8(2) = 16 A3 = 16(2) = 32 A4 = 32(2) = 64 A5 = 64(2) = 128 A6 = 128(2)= 256 A7 = 512 So, the five geometric means between 8 and 512 are 16, 32, 64, 128, 256.
  • 7.
    Example 3. Findthe geometric mean between 5 and 8. Steps Solution 1. Identify the given information Hence m = 5 and n = 8 2. Solve the geometric mean using the formula of 𝑚𝑛 . 𝑚𝑛 = 5. 8 = 40 = 4. 10 = 2 10 So the geometric mean between 5 and 8 is 2 10.