The coefficients of three consecutive terms of `(1+x)^(n+5)` are in the ratio `5:10:14`. Then, n is equal to :

Answer

To solve the problem, we need to find the value of \( n \) given that the coefficients of three consecutive terms of \( (1+x)^{n+5} \) are in the ratio \( 5:10:14 \).

### Step 1: Identify the Coefficients
The coefficients of three consecutive terms in the expansion of \( (1+x)^{n+5} \) can be represented as:
- \( T_{r-1} = \binom{n+5}{r-1} \)
- \( T_{r} = \binom{n+5}{r} \)
- \( T_{r+1} = \binom{n+5}{r+1} \)

### Step 2: Set Up the Ratios
According to the problem, the ratios of these coefficients are given as:
\[ \frac{T_{r-1}}{T_{r}} = \frac{5}{10} = \frac{1}{2} \]
\[ \frac{T_{r}}{T_{r+1}} = \frac{10}{14} = \frac{5}{7} \]

### Step 3: Write the Equations
From the first ratio:
\[ \frac{\binom{n+5}{r-1}}{\binom{n+5}{r}} = \frac{1}{2} \]
Using the property of binomial coefficients:
\[ \frac{\binom{n+5}{r-1}}{\binom{n+5}{r}} = \frac{r}{n+5-r+1} = \frac{r}{n+6-r} \]
Setting this equal to \( \frac{1}{2} \):
\[ \frac{r}{n+6-r} = \frac{1}{2} \]
Cross-multiplying gives:
\[ 2r = n + 6 - r \]
Rearranging gives:
\[ 3r = n + 6 \quad \text{(Equation 1)} \]

From the second ratio:
\[ \frac{\binom{n+5}{r}}{\binom{n+5}{r+1}} = \frac{5}{7} \]
Using the property of binomial coefficients:
\[ \frac{\binom{n+5}{r}}{\binom{n+5}{r+1}} = \frac{n+5-r}{r+1} \]
Setting this equal to \( \frac{5}{7} \):
\[ \frac{n+5-r}{r+1} = \frac{5}{7} \]
Cross-multiplying gives:
\[ 7(n + 5 - r) = 5(r + 1) \]
Expanding and rearranging gives:
\[ 7n + 35 - 7r = 5r + 5 \]
Combining like terms results in:
\[ 7n + 30 = 12r \quad \text{(Equation 2)} \]

### Step 4: Solve the Equations
Now we have two equations:
1. \( 3r = n + 6 \)
2. \( 7n + 30 = 12r \)

From Equation 1, we can express \( r \) in terms of \( n \):
\[ r = \frac{n + 6}{3} \]

Substituting \( r \) into Equation 2:
\[ 7n + 30 = 12\left(\frac{n + 6}{3}\right) \]
Simplifying gives:
\[ 7n + 30 = 4(n + 6) \]
Expanding and rearranging:
\[ 7n + 30 = 4n + 24 \]
\[ 3n = -6 \]
\[ n = -2 \]

### Conclusion
Thus, the value of \( n \) is:
\[ \boxed{-2} \]