Question

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

Answer 1

$a_2=14$ and $a_3=18$

Common difference, $d=a_3-a_2=18-14=4$

Now

$a_2=a+d=14$

$a+4=14$

$a=10$

Now, sum of 51 terms

$S_n=\frac{n}{2}(2a+(n-1\left)d\right)$

$S_{51}=\frac{51}{2}\left(2\right(10)+(51-1\left)4\right)$

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

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

$=51\times 110=5610$

$S_{51}=5610$