Six positive numbers are in G.P., such that their product is 1000. If the fourth term is 1, then the last term is

Options

A. 1000
B. 100
C. `1//100`
D. `1//1000`

Answer

To solve the problem step by step, we will use the properties of geometric progression (G.P.) and the information given in the question.

### Step-by-Step Solution:

1. **Define the terms in G.P.**:
Let the six positive numbers in G.P. be:
- First term: \( a \)
- Second term: \( ar \)
- Third term: \( ar^2 \)
- Fourth term: \( ar^3 \)
- Fifth term: \( ar^4 \)
- Sixth term: \( ar^5 \)

2. **Use the product of the terms**:
The product of these six terms is given to be 1000. Therefore:
\[ a \cdot ar \cdot ar^2 \cdot ar^3 \cdot ar^4 \cdot ar^5 = 1000 \]
This simplifies to:
\[ a^6 \cdot r^{1+2+3+4+5} = 1000 \]
The sum of the powers of \( r \) is:
\[ 1 + 2 + 3 + 4 + 5 = 15 \]
Thus, we have:
\[ a^6 \cdot r^{15} = 1000 \]

3. **Express 1000 in terms of powers**:
We can express 1000 as:
\[ 1000 = 10^3 \]
Therefore, we have:
\[ a^6 \cdot r^{15} = 10^3 \]

4. **Use the information about the fourth term**:
We are given that the fourth term is 1:
\[ ar^3 = 1 \]

5. **Express \( a \) in terms of \( r \)**:
From the fourth term equation, we can express \( a \):
\[ a = \frac{1}{r^3} \]

6. **Substitute \( a \) in the product equation**:
Substitute \( a = \frac{1}{r^3} \) into the product equation:
\[ \left(\frac{1}{r^3}\right)^6 \cdot r^{15} = 10^3 \]
This simplifies to:
\[ \frac{1}{r^{18}} \cdot r^{15} = 10^3 \]
Which gives:
\[ \frac{1}{r^3} = 10^3 \]
Therefore:
\[ r^3 = \frac{1}{1000} \]
Taking the cube root:
\[ r = \frac{1}{10} \]

7. **Find \( a \)**:
Substitute \( r \) back to find \( a \):
\[ a = \frac{1}{\left(\frac{1}{10}\right)^3} = 10^3 = 1000 \]

8. **Find the last term**:
The last term (sixth term) is:
\[ ar^5 = 1000 \cdot \left(\frac{1}{10}\right)^5 = 1000 \cdot \frac{1}{100000} = \frac{1000}{100000} = \frac{1}{100} \]

### Final Answer:
The last term is \( \frac{1}{100} \).