Find the seventh term of the G.P. :
`sqrt(3)+1,1,(sqrt(3)-1)/(2), . . . . .. . . . . `

Answer

To find the seventh term of the geometric progression (G.P.) given by the sequence:

\[ \sqrt{3} + 1, 1, \frac{\sqrt{3} - 1}{2}, \ldots \]

we will follow these steps:

### Step 1: Identify the first term and the second term
The first term \( a_1 \) is given as:
\[ a_1 = \sqrt{3} + 1 \]
The second term \( a_2 \) is:
\[ a_2 = 1 \]

### Step 2: Calculate the common ratio \( r \)
The common ratio \( r \) can be calculated using the formula:
\[ r = \frac{a_2}{a_1} \]
Substituting the values we have:
\[ r = \frac{1}{\sqrt{3} + 1} \]

### Step 3: Rationalize the common ratio
To simplify \( r \), we can rationalize the denominator:
\[ r = \frac{1}{\sqrt{3} + 1} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{\sqrt{3} - 1}{(\sqrt{3})^2 - (1)^2} = \frac{\sqrt{3} - 1}{3 - 1} = \frac{\sqrt{3} - 1}{2} \]

### Step 4: Use the formula for the \( n \)-th term of a G.P.
The formula for the \( n \)-th term of a G.P. is:
\[ a_n = a_1 \cdot r^{n-1} \]
For the seventh term \( a_7 \):
\[ a_7 = a_1 \cdot r^{7-1} = a_1 \cdot r^6 \]

### Step 5: Substitute the values into the formula
Substituting \( a_1 \) and \( r \):
\[ a_7 = (\sqrt{3} + 1) \cdot \left(\frac{\sqrt{3} - 1}{2}\right)^6 \]

### Step 6: Simplify the expression
Calculating \( \left(\frac{\sqrt{3} - 1}{2}\right)^6 \):
\[ \left(\frac{\sqrt{3} - 1}{2}\right)^6 = \frac{(\sqrt{3} - 1)^6}{2^6} = \frac{(\sqrt{3} - 1)^6}{64} \]

Thus,
\[ a_7 = (\sqrt{3} + 1) \cdot \frac{(\sqrt{3} - 1)^6}{64} \]

### Step 7: Final expression
The final expression for the seventh term is:
\[ a_7 = \frac{(\sqrt{3} + 1)(\sqrt{3} - 1)^6}{64} \]