If the first three terms in the expansion of `(1+ax)^n ` are `1,12 x , 64x^2 , ` find 'n' and 'a'

Answer

To solve the problem, we need to find the values of \( n \) and \( a \) given the first three terms of the expansion of \( (1 + ax)^n \) are \( 1, 12x, \) and \( 64x^2 \).

### Step 1: Write the first three terms of the binomial expansion
The binomial expansion of \( (1 + ax)^n \) can be expressed as:
\[ T_k = \binom{n}{k} (ax)^k \]
where \( T_k \) is the \( k^{th} \) term in the expansion.

The first three terms are:
- \( T_1 = \binom{n}{0} (ax)^0 = 1 \)
- \( T_2 = \binom{n}{1} (ax)^1 = n \cdot ax \)
- \( T_3 = \binom{n}{2} (ax)^2 = \frac{n(n-1)}{2} (ax)^2 \)

### Step 2: Set up equations based on the given terms
From the problem, we know:
- \( T_1 = 1 \)
- \( T_2 = 12x \)
- \( T_3 = 64x^2 \)

From \( T_2 \):
\[ n \cdot a = 12 \quad \text{(1)} \]

From \( T_3 \):
\[ \frac{n(n-1)}{2} a^2 = 64 \quad \text{(2)} \]

### Step 3: Solve the equations
From equation (1), we can express \( a \) in terms of \( n \):
\[ a = \frac{12}{n} \quad \text{(3)} \]

Substituting equation (3) into equation (2):
\[ \frac{n(n-1)}{2} \left(\frac{12}{n}\right)^2 = 64 \]

### Step 4: Simplify the equation
This simplifies to:
\[ \frac{n(n-1)}{2} \cdot \frac{144}{n^2} = 64 \]
\[ \frac{72(n-1)}{n} = 64 \]

### Step 5: Cross-multiply and solve for \( n \)
Cross-multiplying gives:
\[ 72(n - 1) = 64n \]
\[ 72n - 72 = 64n \]
\[ 72n - 64n = 72 \]
\[ 8n = 72 \]
\[ n = 9 \]

### Step 6: Substitute \( n \) back to find \( a \)
Using equation (3):
\[ a = \frac{12}{n} = \frac{12}{9} = \frac{4}{3} \]

### Final Answer
Thus, the values of \( n \) and \( a \) are:
\[ n = 9, \quad a = \frac{4}{3} \]