Find the next three terms of the sequence :
`sqrt(5),5,5sqrt(5), . . . . . . . . .. `

Answer

To find the next three terms of the sequence: \( \sqrt{5}, 5, 5\sqrt{5}, \ldots \), we will first identify if the sequence is a geometric progression (GP) and then calculate the next terms.

### Step 1: Identify the first term and the common ratio
The first term \( a \) of the sequence is:
\[ a = \sqrt{5} \]

The second term is:
\[ t_2 = 5 \]

To find the common ratio \( r \), we divide the second term by the first term:
\[ r = \frac{t_2}{t_1} = \frac{5}{\sqrt{5}} = \sqrt{5} \]

### Step 2: Verify the third term
Now let's verify the third term \( t_3 \) using the common ratio:
\[ t_3 = a \cdot r^2 = \sqrt{5} \cdot (\sqrt{5})^2 = \sqrt{5} \cdot 5 = 5\sqrt{5} \]
This confirms that the sequence is indeed a GP.

### Step 3: Calculate the next three terms
We will now calculate the fourth, fifth, and sixth terms using the formula for the \( n \)-th term of a GP:
\[ t_n = a \cdot r^{n-1} \]

#### Fourth Term \( t_4 \):
\[ t_4 = a \cdot r^{4-1} = \sqrt{5} \cdot (\sqrt{5})^3 = \sqrt{5} \cdot 5\sqrt{5} = 5 \cdot 5 = 25 \]

#### Fifth Term \( t_5 \):
\[ t_5 = a \cdot r^{5-1} = \sqrt{5} \cdot (\sqrt{5})^4 = \sqrt{5} \cdot 25 = 25\sqrt{5} \]

#### Sixth Term \( t_6 \):
\[ t_6 = a \cdot r^{6-1} = \sqrt{5} \cdot (\sqrt{5})^5 = \sqrt{5} \cdot 125 = 125 \]

### Conclusion
The next three terms of the sequence are:
1. \( 25 \)
2. \( 25\sqrt{5} \)
3. \( 125 \)