Call the row $x=t+k$, and the position of the terms $t-1, t, t+1$. Call the middle term in the ratio $N = \dbinom{t+k}{t} = \frac{(t+k)!}{k!t!}$. The first term is $N \frac{t}{k+1}$, and the final term is $N \frac{k}{t+1}$. Because we have the ratio $3:4:5$,
$\frac{t}{k+1} = \frac{3}{4}$ and $\frac{k}{t+1} = \frac{5}{4}$.
$4t = 3k+3$ and $4k= 5t+5$
$4t-3k=3$
$5t-4k=-5$
Solve the equations to get $t= 27, k=35$ and $x = t+k = \boxed{062}$.
-Solution and LaTeX by jackshi2006, variables and algebra simplified by oinava




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In Pascal's Triangle, each entry is the sum of the two entries above it. The first few rows of the triangle are shown below.
\[\begin{array}{c@{\hspace{8em}} c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{4pt}}c@{\hspace{2pt}} c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{3pt}}c@{\hspace{6pt}} c@{\hspace{6pt}}c@{\hspace{6pt}}c} \vspace{4pt} \text{Row 0: } &    &    &     &     &    &    & 1 &     &     &    &    &    &  \\\vspace{4pt} \text{Row 1: } &    &    &     &     &    & 1 &    & 1  &     &    &    &    &  \\\vspace{4pt} \text{Row 2: } &    &    &     &     & 1 &    & 2 &     & 1  &    &    &    &  \\\vspace{4pt} \text{Row 3: } &    &    &     &  1 &    & 3 &    & 3  &     & 1 &    &    &  \\\vspace{4pt} \text{Row 4: } &    &    & 1  &     & 4 &    & 6 &     & 4  &    & 1 &    &  \\\vspace{4pt} \text{Row 5: } &    & 1 &     & 5  &    &10&    &10 &     & 5 &    & 1 &  \\\vspace{4pt} \text{Row 6: } & 1 &    & 6  &     &15&    &20&     &15 &    & 6 &    & 1 \end{array}\]
In which row of [Pascal's Triangle](https://artofproblemsolving.com/wiki/index.php/Pascal%27s_Triangle) do three consecutive entries occur that are in the ratio $3 :4 :5$?