Find the sum of the first $51$ terms of an AP whose second and third terms are $14$ and $18$ respectively.

Let the first term be $a$ and common difference be $d$. Then $$a+d=14, \quad a+2d=18.$$ Subtracting gives $d=4$, so $a=10$. Now $$S_{51}=\frac{51}{2}[2\cdot 10+(51-1)\cdot 4]=\frac{51}{2}(20+200)=\frac{51}{2}\cdot 220=51\cdot 110=5610.$$ Therefore the sum is $5610$.