In a binomial expansion, `( x+ a)^(n)`, the first three terms are `1, 56 and 1372` respectively. Find values of `x and a`.

Answer

To solve the problem, we need to find the values of \( x \) and \( a \) given the first three terms of the binomial expansion \( (x + a)^n \) are \( 1, 56, \) and \( 1372 \).

### Step-by-Step Solution:

1. **Identify the first term**:
The first term of the binomial expansion is given by:
\[ T_1 = \binom{n}{0} x^n a^0 = x^n \]
We know from the problem that \( T_1 = 1 \). Therefore:
\[ x^n = 1 \quad \text{(Equation 1)} \]

2. **Identify the second term**:
The second term is given by:
\[ T_2 = \binom{n}{1} x^{n-1} a^1 = n x^{n-1} a \]
We know \( T_2 = 56 \). Thus:
\[ n x^{n-1} a = 56 \quad \text{(Equation 2)} \]

3. **Identify the third term**:
The third term is given by:
\[ T_3 = \binom{n}{2} x^{n-2} a^2 = \frac{n(n-1)}{2} x^{n-2} a^2 \]
We know \( T_3 = 1372 \). Therefore:
\[ \frac{n(n-1)}{2} x^{n-2} a^2 = 1372 \quad \text{(Equation 3)} \]

4. **Substituting \( x \) from Equation 1**:
From Equation 1, since \( x^n = 1 \), we can conclude that:
\[ x = 1 \quad \text{(since \( n \) is a positive integer)} \]

5. **Substituting \( x \) into Equations 2 and 3**:
Substitute \( x = 1 \) into Equation 2:
\[ n \cdot 1^{n-1} \cdot a = 56 \implies n a = 56 \quad \text{(Equation 4)} \]
Substitute \( x = 1 \) into Equation 3:
\[ \frac{n(n-1)}{2} \cdot 1^{n-2} \cdot a^2 = 1372 \implies \frac{n(n-1)}{2} a^2 = 1372 \quad \text{(Equation 5)} \]

6. **Express \( n \) in terms of \( a \)**:
From Equation 4, we can express \( n \) as:
\[ n = \frac{56}{a} \quad \text{(Equation 6)} \]

7. **Substituting \( n \) into Equation 5**:
Substitute \( n = \frac{56}{a} \) into Equation 5:
\[ \frac{\frac{56}{a} \left(\frac{56}{a} - 1\right)}{2} a^2 = 1372 \]
Simplifying this gives:
\[ \frac{56(56 - a)}{2a} = 1372 \]
\[ 56(56 - a) = 2744a \]
\[ 3136 - 56a = 2744a \]
\[ 3136 = 2800a \]
\[ a = \frac{3136}{2800} = \frac{392}{350} = 7 \]

8. **Finding \( n \)**:
Substitute \( a = 7 \) back into Equation 4:
\[ n \cdot 7 = 56 \implies n = \frac{56}{7} = 8 \]

9. **Final Values**:
Therefore, the values are:
\[ x = 1 \quad \text{and} \quad a = 7 \]

### Summary:
The values of \( x \) and \( a \) are:
\[ \boxed{x = 1} \quad \text{and} \quad \boxed{a = 7} \]