Question

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

Answer 1

Given that, $a_2 = 14$ and $a_3 = 18$
$d = a_3 − a_2 = 18 − 14 = 4$
$a_2 = a + d$
$14 = a + 4$
$a = 10$
$Sn = \frac n2[2a + (n-1)d]$

$S_{51} = \frac {51}{2} [2 \times 10 + (50-1)4]$

$S_{51}= \frac {51}{2} [20 + (50)4]$

$S_{51}= \frac {51 \times 220}{2}$
$S_{51}= 51 \times 110$
$S_{51} = 5610$