If the `3^(rd)`, `4^(th)` and `5^(th)` terms in the expansion of `(x + a)^(n)` are 84, 280 and 560 respectively then find the values of a, x and n.

Answer

To solve the problem, we need to find the values of \( a \), \( x \), and \( n \) given the 3rd, 4th, and 5th terms of the binomial expansion of \( (x + a)^n \) are 84, 280, and 560 respectively.

### Step-by-Step Solution:

1. **Identify the General Term**:
The \( r^{th} \) term in the expansion of \( (x + a)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} x^{n-r} a^r \]
Therefore, we can express the 3rd, 4th, and 5th terms as:
- 3rd term: \( T_3 = \binom{n}{2} x^{n-2} a^2 = 84 \)
- 4th term: \( T_4 = \binom{n}{3} x^{n-3} a^3 = 280 \)
- 5th term: \( T_5 = \binom{n}{4} x^{n-4} a^4 = 560 \)

2. **Set Up the Equations**:
We have the following three equations:
\[ \binom{n}{2} x^{n-2} a^2 = 84 \quad (1) \]
\[ \binom{n}{3} x^{n-3} a^3 = 280 \quad (2) \]
\[ \binom{n}{4} x^{n-4} a^4 = 560 \quad (3) \]

3. **Divide Equations to Eliminate Terms**:
Divide equation (2) by equation (1):
\[ \frac{\binom{n}{3} x^{n-3} a^3}{\binom{n}{2} x^{n-2} a^2} = \frac{280}{84} \]
Simplifying gives:
\[ \frac{\binom{n}{3}}{\binom{n}{2}} \cdot \frac{a}{x} = \frac{10}{3} \]
Using the property of binomial coefficients:
\[ \frac{\binom{n}{3}}{\binom{n}{2}} = \frac{n-2}{3} \]
Therefore:
\[ \frac{n-2}{3} \cdot \frac{a}{x} = \frac{10}{3} \]
This simplifies to:
\[ (n-2)a = 10x \quad (4) \]

4. **Divide Equations Again**:
Now divide equation (3) by equation (2):
\[ \frac{\binom{n}{4} x^{n-4} a^4}{\binom{n}{3} x^{n-3} a^3} = \frac{560}{280} \]
Simplifying gives:
\[ \frac{\binom{n}{4}}{\binom{n}{3}} \cdot \frac{a}{x} = 2 \]
Using the property of binomial coefficients:
\[ \frac{\binom{n}{4}}{\binom{n}{3}} = \frac{n-3}{4} \]
Therefore:
\[ \frac{n-3}{4} \cdot \frac{a}{x} = 2 \]
This simplifies to:
\[ (n-3)a = 8x \quad (5) \]

5. **Solve the System of Equations**:
Now we have two equations:
- From (4): \( (n-2)a = 10x \)
- From (5): \( (n-3)a = 8x \)

We can express \( a \) in terms of \( x \) from both equations:
From (4):
\[ a = \frac{10x}{n-2} \]
From (5):
\[ a = \frac{8x}{n-3} \]

Setting these equal to each other:
\[ \frac{10x}{n-2} = \frac{8x}{n-3} \]
Assuming \( x \neq 0 \), we can cancel \( x \):
\[ 10(n-3) = 8(n-2) \]
Expanding gives:
\[ 10n - 30 = 8n - 16 \]
Rearranging gives:
\[ 2n = 14 \implies n = 7 \]

6. **Substituting Back to Find \( a \) and \( x \)**:
Substitute \( n = 7 \) into equation (4):
\[ (7-2)a = 10x \implies 5a = 10x \implies a = 2x \]

Substitute \( a = 2x \) into equation (1):
\[ \binom{7}{2} x^{7-2} (2x)^2 = 84 \]
This simplifies to:
\[ 21x^5 \cdot 4x^2 = 84 \implies 84x^7 = 84 \implies x^7 = 1 \implies x = 1 \]

Now substitute \( x = 1 \) back to find \( a \):
\[ a = 2 \cdot 1 = 2 \]

### Final Values:
- \( a = 2 \)
- \( x = 1 \)
- \( n = 7 \)