Find the sum of the first $51$ terms of an arithmetic progression, whose $2^{nd}$ and $3^{rd}$ terms are $14$ and $18$ respectively.
Let $a = $ first term of A.P; $\;$ $d = $ common ratio
$n^{th}$ term of A.P $= t_n = a + \left(n - 1\right)d$
$2^{nd}$ term of A.P $= t_2 = a + d = 14$ $\;\;\; \cdots \; (1)$
$3^{rd}$ term of A.P $= t_3 = a + 2d = 18$ $\;\;\; \cdots \; (2)$
Subtracting equations $(1)$ and $(2)$ we get, $\;\;\;$ $d = 4$
Substituting the value of $d$ in equation $(1)$ gives $\;\;\;$ $a = 14 - 4 = 10$
Sum of $n$ terms of A.P $= S_n = \dfrac{n}{2} \left[2a + \left(n - 1\right)d\right]$
$\therefore \;$ Sum of first $51$ terms $= S_{51} = \dfrac{51}{2} \left[2 \times 10 + \left(51 - 1\right) \times 4\right]$
i.e. $\;$ $S_{51} = \dfrac{51}{2} \left[20 + 200\right]$
i.e. $\;$ $S_{51} = \dfrac{51}{2} \times 220 = 5610$