In G.P`2sqrt(2,4)`,………`128sqrt(2)`, find the 4 term from the end.

Answer

To find the 4th term from the end of the geometric progression (G.P.) given as \(2\sqrt{2}, 4, \ldots, 128\sqrt{2}\), we will follow these steps:

### Step 1: Identify the first term and the last term
The first term \(a\) of the G.P. is:
\[ a = 2\sqrt{2} \]
The last term \(l\) is:
\[ l = 128\sqrt{2} \]

### Step 2: Find the common ratio
To find the common ratio \(r\), we can use the first two terms:
\[ r = \frac{a_2}{a_1} = \frac{4}{2\sqrt{2}} = \frac{4}{2\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Step 3: Determine the number of terms in the G.P.
We can use the formula for the \(n\)-th term of a G.P.:
\[ l = a \cdot r^{n-1} \]
Substituting the known values:
\[ 128\sqrt{2} = 2\sqrt{2} \cdot (\sqrt{2})^{n-1} \]
Dividing both sides by \(2\sqrt{2}\):
\[ \frac{128\sqrt{2}}{2\sqrt{2}} = (\sqrt{2})^{n-1} \]
This simplifies to:
\[ 64 = (\sqrt{2})^{n-1} \]
Since \(64 = 2^6\), we can express \(64\) in terms of powers of \(2\):
\[ 64 = (2^{1/2})^{n-1} = 2^{(n-1)/2} \]
Setting the exponents equal gives:
\[ 6 = \frac{n-1}{2} \]
Multiplying both sides by \(2\):
\[ 12 = n - 1 \]
Thus,
\[ n = 13 \]

### Step 4: Find the 4th term from the end
The 4th term from the end can be calculated using the formula for the \(n\)-th term from the end:
\[ \text{4th term from the end} = l \cdot r^{-3} \]
Substituting the values:
\[ \text{4th term from the end} = 128\sqrt{2} \cdot (\sqrt{2})^{-3} \]
Calculating \((\sqrt{2})^{-3}\):
\[ (\sqrt{2})^{-3} = \frac{1}{(\sqrt{2})^3} = \frac{1}{2^{3/2}} = \frac{1}{2\sqrt{2}} \]
Now substituting this back:
\[ \text{4th term from the end} = 128\sqrt{2} \cdot \frac{1}{2\sqrt{2}} = \frac{128\sqrt{2}}{2\sqrt{2}} = \frac{128}{2} = 64 \]

### Final Answer
Thus, the 4th term from the end of the G.P. is:
\[ \boxed{64} \]