If for some positive integer $$n$$, the coefficients of three consecutive terms in the binomial expansion of $$(1 + x)^{n+5}$$ are in the ratio 5 : 10 : 14, then the largest coefficient in the expansion is:

First recall the Binomial Theorem. For any non-negative integer $$N$$, the expansion of $$(1+x)^N$$ is

$$ (1+x)^N \;=\; \sum_{r=0}^{N} {N \choose r}\,x^{r}, $$

where the coefficient of the term in $$x^{r}$$ is the binomial coefficient $$\displaystyle {N \choose r} = \frac{N!}{r!\,(N-r)!}.$$

In the given question the power of the binomial is $$N = n+5,$$ so the general coefficient is $$\displaystyle {\,n+5 \choose r}.$$

We are told that for some positive integer $$n$$ three consecutive coefficients are in the ratio $$5:10:14.$$ Let the three consecutive terms correspond to the powers $$r-1,\; r,\; r+1.$$ Therefore we can set

$$ {N \choose r-1} : {N \choose r} : {N \choose r+1}\;=\;5:10:14. $$

Introduce a common factor $$k$$ (so that all three real coefficients are multiples of it) and write

$$ {N \choose r-1}=5k,\qquad {N \choose r}=10k,\qquad {N \choose r+1}=14k. $$

To use the given ratios conveniently we employ the standard relationship between successive binomial coefficients:

$$ \frac{{N \choose r}}{{N \choose r-1}}=\frac{N-r+1}{r},\qquad\text{and}\qquad \frac{{N \choose r}}{{N \choose r+1}}=\frac{r+1}{N-r}. $$

Now compare the first two coefficients. From the given ratio we have

$$ \frac{{N \choose r-1}}{{N \choose r}}=\frac{5}{10}=\frac12. $$

But the theoretical ratio is

$$ \frac{{N \choose r-1}}{{N \choose r}}=\frac{r}{N-r+1}. $$

Equating these two expressions gives

$$ \frac{r}{N-r+1}=\frac12 \;\Longrightarrow\;2r=N-r+1 \;\Longrightarrow\;3r=N+1 \;\Longrightarrow\;N=3r-1. $$

Next compare the second and the third coefficients. From the given ratio we have

$$ \frac{{N \choose r}}{{N \choose r+1}}=\frac{10}{14}=\frac{5}{7}. $$

The theoretical ratio is

$$ \frac{{N \choose r}}{{N \choose r+1}}=\frac{r+1}{N-r}. $$

Equating these expressions yields

$$ \frac{r+1}{N-r}=\frac{5}{7} \;\Longrightarrow\;7(r+1)=5(N-r) \;\Longrightarrow\;7r+7=5N-5r \;\Longrightarrow\;12r+7=5N. $$

Substitute $$N=3r-1$$ from the first relation into the second:

$$ 12r+7=5(3r-1) \;\Longrightarrow\;12r+7=15r-5 \;\Longrightarrow\;3r=12 \;\Longrightarrow\;r=4. $$

With $$r=4,$$ return to $$N=3r-1$$ to find

$$ N=3(4)-1=12-1=11. $$

Remembering that $$N=n+5,$$ we obtain

$$ n+5=11 \;\Longrightarrow\;n=6. $$

We now know the actual expansion is $$(1+x)^{11}.$$ For $$N=11$$ the binomial coefficients are $$\displaystyle {11 \choose 0}, {11 \choose 1},\ldots,{11 \choose 11}.$$ The coefficients increase up to the middle and then decrease symmetrically. Because 11 is odd, the two central (and equal largest) coefficients occur at

$$ r=\frac{11-1}{2}=5 \quad\text{and}\quad r+1=6. $$

Compute one of them:

$$ {11 \choose 5} =\frac{11!}{5!\,6!} =\frac{11\times10\times9\times8\times7}{5\times4\times3\times2\times1} =462. $$

The coefficient $$\displaystyle {11 \choose 6}$$ equals the same value by symmetry, so the largest coefficient in the entire expansion is $$462.$$

Hence, the correct answer is Option A.