For which of the following values of `x ,5t h` term is the numerically greatest term in the expansion of `(1+x//3)^(10)` , a. -2 b. 1.8 c. `2` d. `-1. 9`

Options

A. -2
B. 1.8
C. 2
D. -19

Explanation

To find the value of \( x \) for which the 5th term is the numerically greatest term in the expansion of \( \left(1 + \frac{x}{3}\right)^{10} \), we can follow these steps:

### Step 1: Identify the general term
The general term \( T_r \) in the expansion of \( (a + b)^n \) is given by:
\[ T_r = \binom{n}{r-1} a^{n-(r-1)} b^{r-1} \]
For our case:
- \( a = 1 \)
- \( b = \frac{x}{3} \)
- \( n = 10 \)

Thus, the \( r \)-th term is:
\[ T_r = \binom{10}{r-1} \cdot 1^{10-(r-1)} \cdot \left(\frac{x}{3}\right)^{r-1} = \binom{10}{r-1} \cdot \left(\frac{x}{3}\right)^{r-1} \]

### Step 2: Find the 5th term
The 5th term \( T_5 \) is:
\[ T_5 = \binom{10}{4} \cdot \left(\frac{x}{3}\right)^{4} \]

### Step 3: Set up the conditions for the greatest term
To find when \( T_5 \) is the greatest term, we need to compare it with the 4th term \( T_4 \) and the 6th term \( T_6 \).

#### Condition 1: \( T_5 > T_4 \)
\[ \frac{T_5}{T_4} > 1 \]
Calculating \( T_4 \):
\[ T_4 = \binom{10}{3} \cdot \left(\frac{x}{3}\right)^{3} \]
Thus,
\[ \frac{T_5}{T_4} = \frac{\binom{10}{4} \cdot \left(\frac{x}{3}\right)^{4}}{\binom{10}{3} \cdot \left(\frac{x}{3}\right)^{3}} = \frac{\binom{10}{4}}{\binom{10}{3}} \cdot \frac{x}{3} \]
Using the property of binomial coefficients:
\[ \frac{\binom{10}{4}}{\binom{10}{3}} = \frac{10-3}{4} = \frac{7}{4} \]
So,
\[ \frac{7}{4} \cdot \frac{x}{3} > 1 \implies \frac{7x}{12} > 1 \implies x > \frac{12}{7} \]

#### Condition 2: \( T_6 < T_5 \)
\[ \frac{T_6}{T_5} < 1 \]
Calculating \( T_6 \):
\[ T_6 = \binom{10}{5} \cdot \left(\frac{x}{3}\right)^{5} \]
Thus,
\[ \frac{T_6}{T_5} = \frac{\binom{10}{5} \cdot \left(\frac{x}{3}\right)^{5}}{\binom{10}{4} \cdot \left(\frac{x}{3}\right)^{4}} = \frac{\binom{10}{5}}{\binom{10}{4}} \cdot \frac{x}{3} \]
Using the property of binomial coefficients:
\[ \frac{\binom{10}{5}}{\binom{10}{4}} = \frac{10-4}{5} = \frac{6}{5} \]
So,
\[ \frac{6}{5} \cdot \frac{x}{3} < 1 \implies \frac{6x}{15} < 1 \implies x < \frac{15}{6} = \frac{5}{2} \]

### Step 4: Combine the conditions
From the two conditions, we have:
\[ \frac{12}{7} < x < \frac{5}{2} \]

### Step 5: Evaluate the options
Now we check which of the provided options satisfy this inequality:
- a. \( -2 \) (not valid)
- b. \( 1.8 \) (valid since \( \frac{12}{7} \approx 1.71 < 1.8 < 2.5 \))
- c. \( 2 \) (valid since \( \frac{12}{7} \approx 1.71 < 2 < 2.5 \))
- d. \( -1.9 \) (not valid)

### Conclusion
The values of \( x \) for which the 5th term is the numerically greatest term are \( 1.8 \) and \( 2 \).