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

Let a be the first term and d be the common difference of the AP. Then,

$d=a_{3}-a_{2}=18-14=4$

Now,

$a_{2}=14$                         (Given)

$\Rightarrow a+d=14 \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow a+4=14$

$\Rightarrow a=14-4=10$

Using the formula, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$, we get

$S_{51}=\frac{51}{2}[2 \times 10+(51-1) \times 4]$

$=\frac{51}{2}(20+200)$

$=\frac{51}{2} \times 220$

$=5610$

Hence, the required sum is 5610.