The answer is yes. 

Indeed, by rescaling, without loss of generality (wlog) $c=1$. To simplify the notations, let $f:=F$, $a:=a_n$, $b:=b_n$, $c:=c_n$, $t:=t_n$, $D:=D_n$. Passing to a subsequence, wlog $a\to a_*\in[0,1]$ and $t\to t_*\in(0,1]$. Also, wlog $a+2<c$, since $a\le1$ and $c\to\infty$. Also, wlog $b>c$, since $t>0$ and $c>a$. 

By the convexity of $f$, 
\begin{equation*}
	f(a+1)\ge f(c)+\frac{a+1-c}{b-c}\,(f(b)-f(c)).\tag{1} 
\end{equation*}
Using now the convexity of $f$ again together with the inequality $a+2<c$ and (1), we have 
\begin{align*}
	0\le d&:=\frac{f(a)+f(a+2)}2-f(a+1) \\
	&\le\frac{(c-a-1)f(a)+f(c)}{c-a}-f(a+1) \\ 
	&\le\frac{(c-a-1)f(a)+f(c)+D}{c-a} \\
	&\ \ -\Big(f(c)+\frac{a+1-c}{b-c}\,(f(b)-f(c))\Big) \\
	&=\frac{b-a-1}{b-a}\frac Dt\sim\frac D{t_*}\to0, 
\end{align*}
so that 
\begin{equation*}
	d\to0. \tag{2}
\end{equation*}
On the other hand,
\begin{equation*}
	d\to\frac{f(a_*)+f(a_*+2)}2-f(a_*+1)>0
\end{equation*}
by the strict convexity of $f$. This contradicts (2). $\Box$