2 removed inapplicable result.

To simplify things a little, I describe conditions under which the smallest eigenvalue is strictly positive. These can be adjusted to get equality to zero.

Let $\alpha := a_j=a_k$ be the smallest elements on the diagonal. It can be shown that if \begin{equation*} \frac{2}{n(n-1)}\alpha^2 - \frac{1}{4}\sum_{i < j} (a_i-a_j)^2 < n-1, \end{equation*} then the tridiagonal matrix is positive definite.

If you want necessary

Necessary and sufficient conditions , then things for positive definiteness of the tridiagonal matrix in question are slightly more involveddescribed below.

Definition (Chain Sequence). A sequence $\lbrace x_k \rbrace_{k > 0}$ is a chain sequence if there exists another sequence $\lbrace y_k \rbrace_{k\ge 0}$ such that \begin{equation*} x_k = y_k(1-y_{k-1}), \end{equation*} where $y_0 \in [0,1)$ and $y_k \in (0,1)$ for $k > 0$.

By the Wall-Wetzel Theorem, your tridiagonal matrix is positive definite if and only if

\begin{equation*} \left\lbrace \frac{1}{a_ka_{k+1}} \right\rbrace_{k=1}^{n-1} \end{equation*}

is a chain sequence.

Example. In particular, if the entries of the matrix satisfy,

\begin{equation*} 0 < \frac{1}{a_ka_{k+1}} < \frac{1}{4\cos^2\left(\frac{\pi}{n+1}\right)},\quad k=1,\ldots,n-1, \end{equation*} then it is positive definite.

1. M. Andelic, and C. M. Da Fonesca. Sufficient conditions for positive definiteness of tridiagonal matrices revisited. (2010).
1

To simplify things a little, I describe conditions under which the smallest eigenvalue is strictly positive. These can be adjusted to get equality to zero.

Let $\alpha := a_j=a_k$ be the smallest elements on the diagonal. It can be shown that if \begin{equation*} \frac{2}{n(n-1)}\alpha^2 - \frac{1}{4}\sum_{i < j} (a_i-a_j)^2 < n-1, \end{equation*} then the tridiagonal matrix is positive definite.

If you want necessary and sufficient conditions, then things are slightly more involved.

Definition (Chain Sequence). A sequence $\lbrace x_k \rbrace_{k > 0}$ is a chain sequence if there exists another sequence $\lbrace y_k \rbrace_{k\ge 0}$ such that \begin{equation*} x_k = y_k(1-y_{k-1}), \end{equation*} where $y_0 \in [0,1)$ and $y_k \in (0,1)$ for $k > 0$.

By the Wall-Wetzel Theorem, your tridiagonal matrix is positive definite if and only if

\begin{equation*} \left\lbrace \frac{1}{a_ka_{k+1}} \right\rbrace_{k=1}^{n-1} \end{equation*}

is a chain sequence.