If in the expansion of `(1 + ax)^(n),n in `N, the coefficient of x
and `x^(2) ` are 8 and 24 respectively, then

Options

A. a = 2, n=4
B. a = 4, n=2
C. a = 2, n=6
D. a = -2, n=4

Answer

To solve the problem, we need to find the values of \( a \) and \( n \) given the coefficients of \( x \) and \( x^2 \) in the expansion of \( (1 + ax)^n \).

### Step 1: Identify the General Term
The general term \( T_{r+1} \) in the expansion of \( (1 + ax)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} (ax)^r = \binom{n}{r} a^r x^r \]

### Step 2: Coefficient of \( x \)
The coefficient of \( x \) corresponds to \( r = 1 \):
\[ T_{2} = \binom{n}{1} a^1 = n a \]
According to the problem, this coefficient is equal to 8:
\[ n a = 8 \quad \text{(1)} \]

### Step 3: Coefficient of \( x^2 \)
The coefficient of \( x^2 \) corresponds to \( r = 2 \):
\[ T_{3} = \binom{n}{2} a^2 = \frac{n(n-1)}{2} a^2 \]
According to the problem, this coefficient is equal to 24:
\[ \frac{n(n-1)}{2} a^2 = 24 \quad \text{(2)} \]

### Step 4: Solve the Equations
From equation (1), we have:
\[ n a = 8 \implies a = \frac{8}{n} \]
Substituting \( a \) into equation (2):
\[ \frac{n(n-1)}{2} \left(\frac{8}{n}\right)^2 = 24 \]
Simplifying this:
\[ \frac{n(n-1)}{2} \cdot \frac{64}{n^2} = 24 \]
\[ \frac{32(n-1)}{n} = 24 \]
Multiplying both sides by \( n \):
\[ 32(n-1) = 24n \]
\[ 32n - 32 = 24n \]
\[ 32n - 24n = 32 \]
\[ 8n = 32 \implies n = 4 \]

### Step 5: Find \( a \)
Substituting \( n = 4 \) back into equation (1):
\[ 4a = 8 \implies a = 2 \]

### Final Answer
Thus, the values of \( a \) and \( n \) are:
\[ a = 2, \quad n = 4 \]