Question

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

Answer 1

$second \: term \: is \: 14 \\ a + d = 14 \\ third \: term \: is \: 18 \\ a + 2d = 18 \\ solving \: both \: equation \: we \: get \: a = 10 \: \: d = 4 \\ sum \: of \: 51 \: term \: is \: \frac{n}{2} (2a + (n - 1)d) \\ \frac{51}{2}(2 \times 10 + (51 - 1)4) \\ solving \: we \: get \: 5610$

Answer 2

here is your answer using formula
$s = \frac{n}{2} (2a + (n - 1)d)$