Is there any Constraints?

Let $m = n+5$ and the three coefficients be given by:

$\left( \begin{matrix} m \\ r-1 \end{matrix} \right) : \left( \begin{matrix} m \\ r \end{matrix} \right) : \left( \begin{matrix} m \\ r+1 \end{matrix} \right) = 5:10:14$

$\Rightarrow \dfrac {m!}{(r-1)!(m-r+1)!} : \dfrac {m!}{r!(m-r)!} : \dfrac {m!}{(r+1)!(m-r-1)!} = 5:10:14$

$\Rightarrow \dfrac {(r-1)!(m-r+1)!} {r!(m-r)!} = 2 \quad \Rightarrow \dfrac {m-r+1} {r} = 2 \quad \Rightarrow m = 3r -1 \space ...(1)$

Similarly,

$\Rightarrow \dfrac {r!(m-r)!} {(r+1)!(m-r-1)!} = \dfrac {7}{5} \quad \Rightarrow \dfrac {m-r} {r+1} = \dfrac {7}{5} \quad \Rightarrow 5m = 12r+7 \space ...(2)$

Equation $1 \times 5$ and equate it with equation $2$ :

$5m = 15r - 5 = 12r +7 \quad \Rightarrow r = 4 \quad \Rightarrow m = 11 \quad \Rightarrow n = \boxed{6}$

Let the three consecutive terms whose coefficients are in the ratio 5 : 10 : 14 be the $(r-1)^{th}$ , $(r)^{th}$ and $(r+1)^{th}$ term respectively:

We know that in the expansion of $(1+x)^n$ , the general formula for $r^{th}$ term is:

$T_{r+1}$ = n $C_{r}$ . $x^{r}$

The question here is similar:

Thus,

$T_{(r-2)+1}$ = (n+5) $C_{r-2}$ . $x^{r-2}$ (eq.1)

$T_{(r-1)+1}$ = (n+5) $C_{r-1}$ . $x^{r-1}$ (eq.2)

$T_{r+1}$ = (n+5) $C_{r}$ . $x^{r}$ (eq.3)

From, eq. 1,2 and 3 we get that the coefficients of the three consecutive terms are:

(n+5) $C_{r-2}$ , (n+5) $C_{r-1}$ and (n+5) $C_{r}$ respectively.

taking two coefficients at a time, namely from eq.1 and 2 , eq.2 and 3, we get..

$\frac{(n+5)!}{(r-2)!.(n+5-r+2)}$ / $\frac{(n+5)!}{(r-1)!.(n+5-r+1)}$ = 5/10

$\frac{(n+5)!}{(r-1)!.(n+5-r+1)}$ / $\frac{(n+5)!}{(r)!.(n+5-r)}$ = 10/14

after, solving the two equations we get that :

r= $\frac{(n+9)}{3}$ and 24r = 10n+60

substitute the value of r in latter eq., u will get that $\boxed{n=6}$

The coefficients of $(1+x)^{n+5}$ are given by the (n+5)th row in Pascal's Triangle. Consider how the nth row is constructed - Each element can be described in terms of the previous term as follows: start with 1, then multiply by n and divide by 1 to get the next term, then repeat by multiply by 1 less and dividing by 1 more than in the previous step.

$t_0 = \binom{n}{0} = 1 \\ t_1 = \binom{n}{1} = t_0 \cdot \frac{n}{1} \\ t_2 = \binom{n}{2} = t_1 \cdot \frac{n-1}{2} \\ etc ...$

For the 3 terms of interest, let j and k be the multiplying and dividing factors used to get the second term from the first. Using this method, this means j-1 and k+1 are the multiplying and dividing factors used to get the third term from the second. Therefore,

$\frac{j}{k} = \frac{10}{5} = 2 \Rightarrow j = 2k \\ \frac{j-1}{k+1} = \frac{14}{10} = \frac{7}{5} \Rightarrow 5j - 5 = 7k + 7$

Substitute the first equation into the second to get k = 4, j = 8

Reversing the method above, we get $t_1$ = 11. This means we're in the 11th row of Pascal's Triangle $(1+x)^{n+5} = (1+x)^{11} \Rightarrow n+5=11 \Rightarrow \boxed{n=6}$

In summary, use a little intuition and get some easier equations to solve for.