In the expansion of `(1/2 + 2x/3)^n` when `x=-1/2` ,it is known that `3^(rd)` term is the greatest term Find the possible integral values of n .

Answer

To solve the problem step by step, we will follow the instructions provided in the video transcript and derive the solution systematically.

### Step 1: Substitute \( x = -\frac{1}{2} \) into the expression
We start with the expression:
\[ \left(\frac{1}{2} + \frac{2x}{3}\right)^n \]
Substituting \( x = -\frac{1}{2} \):
\[ \left(\frac{1}{2} + \frac{2(-\frac{1}{2})}{3}\right)^n = \left(\frac{1}{2} - \frac{1}{3}\right)^n \]
Calculating \( \frac{1}{2} - \frac{1}{3} \):
\[ \frac{1}{2} = \frac{3}{6}, \quad \frac{1}{3} = \frac{2}{6} \implies \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} \]
Thus, we have:
\[ \left(\frac{1}{6}\right)^n \]

### Step 2: Identify the \( r \)-th term in the binomial expansion
The \( r \)-th term in the expansion of \( (a + b)^n \) is given by:
\[ T_r = \binom{n}{r-1} a^{n-(r-1)} b^{r-1} \]
In our case, \( a = \frac{1}{2} \) and \( b = -\frac{2}{3} \). For the 3rd term (\( r = 3 \)):
\[ T_3 = \binom{n}{2} \left(\frac{1}{2}\right)^{n-2} \left(-\frac{2}{3}\right)^2 \]
Calculating \( T_3 \):
\[ T_3 = \binom{n}{2} \left(\frac{1}{2}\right)^{n-2} \cdot \frac{4}{9} \]
Thus:
\[ T_3 = \frac{4}{9} \cdot \binom{n}{2} \cdot \left(\frac{1}{2}\right)^{n-2} \]

### Step 3: Determine the condition for the greatest term
For the \( r \)-th term to be the greatest, we use the condition:
\[ r = 1 + \frac{n |x|}{1 + |x|} \]
Here, \( |x| = \frac{1}{2} \), thus:
\[ 3 = 1 + \frac{n \cdot \frac{1}{2}}{1 + \frac{1}{2}} \]
Calculating the right side:
\[ 3 = 1 + \frac{n \cdot \frac{1}{2}}{\frac{3}{2}} \implies 3 = 1 + \frac{n}{3} \]
Subtracting 1 from both sides:
\[ 2 = \frac{n}{3} \implies n = 6 \]

### Step 4: Finding possible integral values of \( n \)
Since the 3rd term is the greatest, we check the terms around it:
- The 2nd term \( T_2 \) must be less than \( T_3 \)
- The 4th term \( T_4 \) must also be less than \( T_3 \)

From the condition derived, we can find that \( n \) can take values such that \( T_3 \) is greater than both \( T_2 \) and \( T_4 \). This leads us to check values of \( n \):
- \( n = 4 \) gives \( T_3 > T_2 \)
- \( n = 5 \) gives \( T_3 > T_2 \)
- \( n = 6 \) gives \( T_3 > T_2 \)

Thus, the possible integral values of \( n \) are \( 4, 5, 6 \).

### Step 5: Calculate the sum of possible values of \( n \)
Adding the possible values:
\[ 4 + 5 + 6 = 15 \]

### Final Answer
The possible integral values of \( n \) are \( 4, 5, 6 \) and their sum is \( 15 \).