We know $$(1+x)^{n+5}=\sum_\limits{r=0}^{n+5}{ }^{n+5} \mathrm{C}_r \cdot x^r$$

Given, the coefficients of three consecutive terms of $$(1+x)^{n+5}$$ are in the ratio $$5: 10: 14$$.

$$\begin{aligned} & \Rightarrow{ }^{n+5} \mathrm{C}_r:{ }^{n+5} \mathrm{C}_{r+1}:{ }^{n+5} \mathrm{C}_{r+2}=5: 10: 14 \\ & \Rightarrow \frac{{ }^{n+5} \mathrm{C}_{r+1}}{{ }^{n+5} \mathrm{C}_r}=\frac{10}{5} \text { and } \frac{{ }^{n+5} \mathrm{C}_{r+2}}{{ }^{n+5} \mathrm{C}_{r+1}}=\frac{14}{10} \\ & \Rightarrow{ }^{n+5} \mathrm{C}_{r+1}=2 .{ }^{n+5} \mathrm{C}_r \text { and } 5 .{ }^{n+5} \mathrm{C}_{r+2}=7 .{ }^{n+5} \mathrm{C}_{r+1} \end{aligned}$$

$$ \Rightarrow {{\left| \!{\underline {\, {n + 5} \,}} \right. } \over {\left| \!{\underline {\, {r + 1} \,}} \right. \,\left| \!{\underline {\, {n - r + 4} \,}} \right. }} = 2{{\left| \!{\underline {\, {n + 5} \,}} \right. } \over {\left| \!{\underline {\, r \,}} \right. \,\left| \!{\underline {\, {n - r + 5} \,}} \right. }}$$ and

$$\,\,\,\,\,\,\,{{5\left| \!{\underline {\, {n + 5} \,}} \right. } \over {\left| \!{\underline {\, {r + 2} \,}} \right. \,\left| \!{\underline {\, {n - r + 3} \,}} \right. }} = {{7\left| \!{\underline {\, {n + 5} \,}} \right. } \over {\left| \!{\underline {\, {r + 1} \,}} \right. \,\left| \!{\underline {\, {n - r + 4} \,}} \right. }}$$

$$\begin{aligned} \Rightarrow \quad & \frac{1}{r+1}=\frac{2}{n-r+5} \text { and } \frac{5}{r+2}=\frac{7}{n-r+4} \\ \Rightarrow \quad & n-r+5=2 r+2 \text { and } \\ & 5 n-5 r+20=7 r+14 \\ \Rightarrow \quad & n=3 r-3=\frac{12 r-6}{5} \\ \Rightarrow \quad & n=3(r-1) \text { and } 15 r-15=12 r-6 \\ \Rightarrow \quad & r=3 \text { and } n=6 \\ \Rightarrow \quad & n=6 \end{aligned}$$

Hints :

(i) Recall $$(1+x)^m=\sum_\limits{r=1}^m{ }^m C_r \cdot x^r$$

(ii) The coefficients of three consecutive terms $$(1+x)^m$$ are $${ }^m \mathrm{C}_r,{ }^m \mathrm{C}_{r+1},{ }^m \mathrm{C}_{r+2}$$.