Answer :
Certainly! Let's tackle the problem step-by-step.
Given:
- 4 geometric means are placed between 5 and [tex]\( x \)[/tex].
- The third geometric mean is 40.
Let's denote the geometric means by [tex]\( G_1, G_2, G_3, G_4 \)[/tex].
### Step 1: Understanding the Geometric Progression
In a geometric progression (GP), each term is obtained by multiplying the previous term by a common ratio [tex]\( r \)[/tex]. Therefore, if we start with 5, the sequence will be:
[tex]\[ 5, G_1, G_2, G_3, G_4, x \][/tex]
### Step 2: Expressing the Means in Terms of the Common Ratio [tex]\( r \)[/tex]
We can express [tex]\( G_1, G_2, G_3, G_4 \)[/tex] as follows:
[tex]\[ G_1 = 5r \][/tex]
[tex]\[ G_2 = 5r^2 \][/tex]
[tex]\[ G_3 = 5r^3 \][/tex]
[tex]\[ G_4 = 5r^4 \][/tex]
### Step 3: Using the Given Information
We are given that the third mean [tex]\( G_3 \)[/tex] is 40:
[tex]\[ 5r^3 = 40 \][/tex]
### Step 4: Solving for [tex]\( r \)[/tex]
Divide both sides by 5:
[tex]\[ r^3 = \frac{40}{5} = 8 \][/tex]
Taking the cube root of both sides:
[tex]\[ r = \sqrt[3]{8} = 2 \][/tex]
### Step 5: Finding the Other Means
Using [tex]\( r = 2 \)[/tex], we can find the other means:
[tex]\[ G_1 = 5r = 5 \cdot 2 = 10 \][/tex]
[tex]\[ G_2 = 5r^2 = 5 \cdot 2^2 = 5 \cdot 4 = 20 \][/tex]
[tex]\[ G_3 = 5r^3 = 40 \quad \text{(already given and we verified it)} \][/tex]
[tex]\[ G_4 = 5r^4 = 5 \cdot 2^4 = 5 \cdot 16 = 80 \][/tex]
### Step 6: Finding [tex]\( x \)[/tex]
Using the common ratio again:
[tex]\[ x = 5r^5 = 5 \cdot 2^5 = 5 \cdot 32 = 160 \][/tex]
### Conclusion
The four geometric means between 5 and [tex]\( x \)[/tex] are 10, 20, 40, and 80. The value of [tex]\( x \)[/tex] is 160.
Given:
- 4 geometric means are placed between 5 and [tex]\( x \)[/tex].
- The third geometric mean is 40.
Let's denote the geometric means by [tex]\( G_1, G_2, G_3, G_4 \)[/tex].
### Step 1: Understanding the Geometric Progression
In a geometric progression (GP), each term is obtained by multiplying the previous term by a common ratio [tex]\( r \)[/tex]. Therefore, if we start with 5, the sequence will be:
[tex]\[ 5, G_1, G_2, G_3, G_4, x \][/tex]
### Step 2: Expressing the Means in Terms of the Common Ratio [tex]\( r \)[/tex]
We can express [tex]\( G_1, G_2, G_3, G_4 \)[/tex] as follows:
[tex]\[ G_1 = 5r \][/tex]
[tex]\[ G_2 = 5r^2 \][/tex]
[tex]\[ G_3 = 5r^3 \][/tex]
[tex]\[ G_4 = 5r^4 \][/tex]
### Step 3: Using the Given Information
We are given that the third mean [tex]\( G_3 \)[/tex] is 40:
[tex]\[ 5r^3 = 40 \][/tex]
### Step 4: Solving for [tex]\( r \)[/tex]
Divide both sides by 5:
[tex]\[ r^3 = \frac{40}{5} = 8 \][/tex]
Taking the cube root of both sides:
[tex]\[ r = \sqrt[3]{8} = 2 \][/tex]
### Step 5: Finding the Other Means
Using [tex]\( r = 2 \)[/tex], we can find the other means:
[tex]\[ G_1 = 5r = 5 \cdot 2 = 10 \][/tex]
[tex]\[ G_2 = 5r^2 = 5 \cdot 2^2 = 5 \cdot 4 = 20 \][/tex]
[tex]\[ G_3 = 5r^3 = 40 \quad \text{(already given and we verified it)} \][/tex]
[tex]\[ G_4 = 5r^4 = 5 \cdot 2^4 = 5 \cdot 16 = 80 \][/tex]
### Step 6: Finding [tex]\( x \)[/tex]
Using the common ratio again:
[tex]\[ x = 5r^5 = 5 \cdot 2^5 = 5 \cdot 32 = 160 \][/tex]
### Conclusion
The four geometric means between 5 and [tex]\( x \)[/tex] are 10, 20, 40, and 80. The value of [tex]\( x \)[/tex] is 160.